{"id":127328,"date":"2021-09-14T14:21:26","date_gmt":"2021-09-14T08:51:26","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=127328"},"modified":"2021-09-14T14:21:32","modified_gmt":"2021-09-14T08:51:32","slug":"rd-sharma-class-10-solutions-chapter-9-exercise-9-6","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-9-exercise-9-6\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions Exercise 9.6 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-127363\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-9-Exercise-9.6.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-9-Exercise-9.6.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-9-Exercise-9.6-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6:\u00a0<\/strong>This exercise is about problems with an A.P.&#8217;s sum of terms. <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> is the best material for students to refer to and study effectively for their exams because it covers a wide range of real problems. The following are the\u00a0<a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-9-arithmetic-progressions\/\"><strong>RD Sharma Class 10 Solutions Chapter 9<\/strong><\/a> Exercise 9.6 PDF exercise solutions.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d91817ee6a5\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path 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Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-9-Exercise-9.6.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-9-Exercise-9.6.pdf\">RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-class-10-solutions-chapter-9-exercise-96-important-question-with-answers\"><\/span>Access answers to RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q2. <\/strong><\/p>\n<p><strong>Find the sum to n terms of the A.P. 5, 2, \u20131, \u2013 4, \u20137, \u2026<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given AP is 5, 2, -1, -4, -7, \u2026..<\/p>\n<p>Here, a = 5, d = 2 \u2013 5 = -3<\/p>\n<p>We know that,<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2{2a + (n \u2013 1)d}<\/p>\n<p>= n\/2{2.5 + (n \u2013 1) \u2013 3}<\/p>\n<p>= n\/2{10 \u2013 3(n \u2013 1)}<\/p>\n<p>= n\/2{13 \u2013 3n)<\/p>\n<p>\u2234 S<sub>n<\/sub>\u00a0= n\/2(13 \u2013 3n)<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q3. <\/strong><\/p>\n<p><strong>Find the sum of n terms of an A.P. whose the terms is given by a<sub>n<\/sub>\u00a0= 5 \u2013 6n.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given nth term of the A.P as a<sub>n<\/sub>\u00a0= 5 \u2013 6n<\/p>\n<p>Put n = 1, we get<\/p>\n<p>a<sub>1<\/sub>\u00a0= 5 \u2013 6.1 = -1<\/p>\n<p>So, first term (a) = -1<\/p>\n<p>Last term (a<sub>n<\/sub>) = 5 \u2013 6n = 1<\/p>\n<p>Then, S<sub>n<\/sub>\u00a0= n\/2(-1 + 5 \u2013 6n)<\/p>\n<p>= n\/2(4 \u2013 6n) = n(2 \u2013 3n)<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q4. <\/strong><\/p>\n<p><strong>Find the sum of last ten terms of the A.P. : 8, 10, 12, 14, .. , 126<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given A.P. 8, 10, 12, 14, .. , 126<\/p>\n<p>Here, a = 8 , d = 10 \u2013 8 = 2<\/p>\n<p>We know that, a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>So, to find the number of terms<\/p>\n<p>126 = 8 + (n \u2013 1)2<\/p>\n<p>126 = 8 + 2n \u2013 2<\/p>\n<p>2n = 120<\/p>\n<p>n = 60<\/p>\n<p>Next, let\u2019s find the 51<sup>st<\/sup>\u00a0term<\/p>\n<p>a<sub>51<\/sub>\u00a0= 8 + 50(2) = 108<\/p>\n<p>So, the sum of last ten terms is the sum of a<sub>51<\/sub>\u00a0+ a<sub>52<\/sub>\u00a0+ a<sub>53<\/sub>\u00a0+ \u2026\u2026. + a<sub>60<\/sub><\/p>\n<p>Here, n = 10, a = 108 and l = 126<\/p>\n<p>S = 10\/2 [108 + 126]<\/p>\n<p>= 5(234)<\/p>\n<p>= 1170<\/p>\n<p>Hence, the sum of the last ten terms of the A.P is 1170.<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q5. <\/strong><\/p>\n<p><strong>Find the sum of the first 15 terms of each of the following sequences having n<sup>th<\/sup>\u00a0term as:<\/strong><\/p>\n<p><strong>(i) a<sub>n<\/sub>\u00a0= 3 + 4n\u00a0<\/strong><\/p>\n<p><strong>(ii) b<sub>n<\/sub>\u00a0= 5 + 2n\u00a0<\/strong><\/p>\n<p><strong>(iii) x<sub>n<\/sub>\u00a0= 6 \u2013 n\u00a0<\/strong><\/p>\n<p><strong>(iv) y<sub>n<\/sub>\u00a0= 9 \u2013 5n<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Given an A.P. whose n<sup>th<\/sup>\u00a0term is given by a<sub>n<\/sub>\u00a0= 3 + 4n<\/p>\n<p>To find the sum of the n terms of the given A.P., using the formula,<\/p>\n<p>S<sub>n<\/sub>\u00a0= n(a + l)\/ 2<\/p>\n<p>Where, a = the first term l = the last term.<\/p>\n<p>Putting n = 1 in the given a<sub>n<\/sub>, we get<\/p>\n<p>a = 3 + 4(1) = 3 + 4 = 7<\/p>\n<p>For the last term (l), here n = 15<\/p>\n<p>a<sub>15<\/sub>\u00a0= 3 + 4(15) = 63<\/p>\n<p>So,\u00a0S<sub>n<\/sub>\u00a0= 15(7 + 63)\/2<\/p>\n<p>= 15 x 35<\/p>\n<p>= 525<\/p>\n<p>Therefore, the sum of the 15 terms of the given A.P. is S<sub>15<\/sub>\u00a0= 525<\/p>\n<p>(ii) Given an A.P. whose n<sup>th<\/sup>\u00a0term is given by b<sub>n<\/sub>\u00a0= 5 + 2n<\/p>\n<p>To find the sum of the n terms of the given A.P., using the formula,<\/p>\n<p>S<sub>n<\/sub>\u00a0= n(a + l)\/ 2<\/p>\n<p>Where, a = the first term l = the last term.<\/p>\n<p>Putting n = 1 in the given b<sub>n<\/sub>, we get<\/p>\n<p>a = 5 + 2(1) = 5 + 2 = 7<\/p>\n<p>For the last term (l), here n = 15<\/p>\n<p>a<sub>15<\/sub>\u00a0= 5 + 2(15) = 35<\/p>\n<p>So,\u00a0S<sub>n<\/sub>\u00a0= 15(7 + 35)\/2<\/p>\n<p>= 15 x 21<\/p>\n<p>= 315<\/p>\n<p>Therefore, the sum of the 15 terms of the given A.P. is S<sub>15<\/sub>\u00a0= 315<\/p>\n<p>(iii) Given an A.P. whose n<sup>th<\/sup>\u00a0term is given by x<sub>n<\/sub>\u00a0= 6 \u2013 n<\/p>\n<p>To find the sum of the n terms of the given A.P., using the formula<\/p>\n<p>S<sub>n<\/sub>\u00a0= n(a + l)\/ 2<\/p>\n<p>Where, a = the first term l = the last term.<\/p>\n<p>Putting n = 1 in the given x<sub>n<\/sub>, we get<\/p>\n<p>a = 6 \u2013 1 = 5<\/p>\n<p>For the last term (l), here n = 15<\/p>\n<p>a<sub>15<\/sub>\u00a0= 6 \u2013 15 = -9<\/p>\n<p>So,\u00a0S<sub>n<\/sub>\u00a0= 15(5 \u2013 9)\/2<\/p>\n<p>= 15 x (-2)<\/p>\n<p>= -30<\/p>\n<p>Therefore, the sum of the 15 terms of the given A.P. is S<sub>15<\/sub>\u00a0= -30<\/p>\n<p>(iv) Given an A.P. whose n<sup>th<\/sup>\u00a0term is given by y<sub>n<\/sub>\u00a0= 9 \u2013 5n<\/p>\n<p>To find the sum of the n terms of the given A.P., using the formula,<\/p>\n<p>S<sub>n<\/sub>\u00a0= n(a + l)\/ 2<\/p>\n<p>Where, a = the first term l = the last term.<\/p>\n<p>Putting n = 1 in the given y<sub>n<\/sub>, we get<\/p>\n<p>a = 9 \u2013 5(1) = 9 \u2013 5 = 4<\/p>\n<p>For the last term (l), here n = 15<\/p>\n<p>a<sub>15<\/sub>\u00a0= 9 \u2013 5(15) = -66<\/p>\n<p>So,\u00a0S<sub>n<\/sub>\u00a0= 15(4 \u2013 66)\/2<\/p>\n<p>= 15 x (-31)<\/p>\n<p>= -465<\/p>\n<p>Therefore, the sum of the 15 terms of the given A.P. is S<sub>15<\/sub>\u00a0= -465<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q6. <\/strong><\/p>\n<p><strong>Find the sum of the first 20 terms the sequence whose n<sup>th<\/sup>\u00a0term is a<sub>n<\/sub>\u00a0= An + B.<\/strong><\/p>\n<p><strong>Solution:<br \/><\/strong><\/p>\n<p>Given an A.P. whose nth term is given by, a<sub>n<\/sub>\u00a0= An + B<\/p>\n<p>We need to find the sum of the first 20 terms.<\/p>\n<p>To find the sum of the n terms of the given A.P., we use the formula,<\/p>\n<p>S<sub>n<\/sub>\u00a0= n(a + l)\/ 2<\/p>\n<p>Where, a = the first term l = the last term,<\/p>\n<p>Putting n = 1 in the given a<sub>n<\/sub>, we get<\/p>\n<p>a = A(1) + B = A + B<\/p>\n<p>For the last term (l), here n = 20<\/p>\n<p>A<sub>20<\/sub>\u00a0= A(20) + B = 20A + B<\/p>\n<p>S<sub>20<\/sub>\u00a0= 20\/2((A + B) + 20A + B)<\/p>\n<p>= 10[21A + 2B]<\/p>\n<p>= 210A + 20B<\/p>\n<p>Therefore, the sum of the first 20 terms of the given A.P. is 210 A + 20B<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q7. <\/strong><\/p>\n<p><strong>Find the sum of the first 25 terms of an A.P whose n<sup>th<\/sup>\u00a0term is given by a<sub>n<\/sub>\u00a0= 2 \u2013 3n.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given an A.P. whose n<sup>th<\/sup>\u00a0term is given by a<sub>n<\/sub>\u00a0= 2 \u2013 3n<\/p>\n<p>To find the sum of the n terms of the given A.P., we use the formula,<\/p>\n<p>S<sub>n<\/sub>\u00a0= n(a + l)\/ 2<\/p>\n<p>Where, a = the first term l = the last term.<\/p>\n<p>Putting n = 1 in the given a<sub>n<\/sub>, we get<\/p>\n<p>a = 2 \u2013 3(1) = -1<\/p>\n<p>For the last term (l), here n = 25<\/p>\n<p>a<sub>25<\/sub>\u00a0= 2 \u2013 3(25) = -73<\/p>\n<p>So,\u00a0S<sub>n<\/sub>\u00a0= 25(-1 \u2013 73)\/2<\/p>\n<p>= 25 x (-37)<\/p>\n<p>= -925<\/p>\n<p>Therefore, the sum of the 25 terms of the given A.P. is S<sub>25<\/sub>\u00a0= -925<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q8. <\/strong><\/p>\n<p><strong>Find the sum of the first 25 terms of an A.P whose n<sup>th<\/sup>\u00a0term is given by a<sub>n<\/sub>\u00a0= 7 \u2013 3n.\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given an A.P. whose n<sup>th<\/sup>\u00a0term is given by a<sub>n<\/sub>\u00a0= 7 \u2013 3n<\/p>\n<p>To find the sum of the n terms of the given A.P., we use the formula,<\/p>\n<p>S<sub>n<\/sub>\u00a0= n(a + l)\/ 2<\/p>\n<p>Where, a = the first term l = the last term.<\/p>\n<p>Putting n = 1 in the given a<sub>n<\/sub>, we get<\/p>\n<p>a = 7 \u2013 3(1) = 7 \u2013 3 = 4<\/p>\n<p>For the last term (l), here n = 25<\/p>\n<p>a<sub>15<\/sub>\u00a0= 7 \u2013 3(25) = -68<\/p>\n<p>So,\u00a0S<sub>n<\/sub>\u00a0= 25(4 \u2013 68)\/2<\/p>\n<p>= 25 x (-32)<\/p>\n<p>= -800<\/p>\n<p>Therefore, the sum of the 15 terms of the given A.P. is S<sub>25<\/sub>\u00a0= -800<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q9. <\/strong><\/p>\n<p><strong>If the sum of a certain number of terms starting from the first term of an A.P. is 25, 22, 19, . . ., is 116. Find the last term.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given the sum of the certain number of terms of an A.P. = 116<\/p>\n<p>We know that, S<sub>n<\/sub>\u00a0= n\/2[2a + (n \u2212 1)d]<\/p>\n<p>Where; a = first term for the given A.P.<\/p>\n<p>d = common difference of the given A.P.<\/p>\n<p>n = number of terms So for the given A.P.(25, 22, 19,\u2026)<\/p>\n<p>Here we have, the first term (a) = 25<\/p>\n<p>The sum of n terms S<sub>n<\/sub>\u00a0= 116<\/p>\n<p>Common difference of the A.P. (d) = a<sub>2<\/sub>\u00a0\u2013 a<sub>1<\/sub>\u00a0= 22 \u2013 25 = -3<\/p>\n<p>Now, substituting values in S<sub>n<\/sub><\/p>\n<p>\u27f9\u00a0116 = n\/2[2(25) + (n \u2212 1)(\u22123)]<\/p>\n<p>\u27f9\u00a0(n\/2)[50 + (\u22123n + 3)] = 116<\/p>\n<p>\u27f9\u00a0(n\/2)[53 \u2212 3n]\u00a0= 116<\/p>\n<p>\u27f9 53n \u2013 3n<sup>2<\/sup>\u00a0= 116 x 2<\/p>\n<p>Thus, we get the following quadratic equation,<\/p>\n<p>3n<sup>2<\/sup>\u00a0\u2013 53n + 232 = 0<\/p>\n<p>By factorization method of solving, we have<\/p>\n<p>\u27f9 3n<sup>2<\/sup>\u00a0\u2013 24n \u2013 29n + 232 = 0<\/p>\n<p>\u27f9 3n( n \u2013 8 ) \u2013 29 ( n \u2013 8 ) = 0<\/p>\n<p>\u27f9 (3n \u2013 29)( n \u2013 8 ) = 0<\/p>\n<p>So, 3n \u2013 29 = 0<\/p>\n<p>\u27f9 n =\u00a029\/3<\/p>\n<p>Also, n \u2013 8 = 0<\/p>\n<p>\u27f9 n = 8<\/p>\n<p>Since n cannot be a fraction, so the number of terms is taken as 8.<\/p>\n<p>So, the term is:<\/p>\n<p>a<sub>8<\/sub>\u00a0= a<sub>1<\/sub>\u00a0+ 7d = 25 + 7(-3) = 25 \u2013 21 =\u00a04<\/p>\n<p>Hence, the last term of the given A.P. such that the sum of the terms is 116 is\u00a04.<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q10. <\/strong><\/p>\n<p><strong>(i) How many terms of the sequence 18, 16, 14\u2026. should be taken so that their sum is zero.\u00a0<\/strong><\/p>\n<p><strong>(ii) How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40?\u00a0<\/strong><\/p>\n<p><strong>(iii) How many terms of the A.P. 9, 17, 25, . . . must be taken so that their sum is 636?\u00a0<\/strong><\/p>\n<p><strong>(iv) How many terms of the A.P. 63, 60, 57, . . . must be taken so that their sum is 693?\u00a0<\/strong><\/p>\n<p><strong>(v) How many terms of the A.P. is 27, 24, 21. . . should be taken that their sum is zero?\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Given AP. is 18, 16, 14, \u2026<\/p>\n<p>We know that,<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2[2a + (n \u2212 1)d]<\/p>\n<p>Here,<\/p>\n<p>The first term (a) = 18<\/p>\n<p>The sum of n terms (S<sub>n<\/sub>) = 0 (given)<\/p>\n<p>Common difference of the A.P.<\/p>\n<p>(d) = a<sub>2<\/sub> \u2013 a<sub>1<\/sub>\u00a0= 16 \u2013 18 = \u2013 2<\/p>\n<p>So, on substituting the values in S<sub>n<\/sub><\/p>\n<p>\u27f9\u00a00 = n\/2[2(18) + (n \u2212 1)(\u22122)]<\/p>\n<p>\u27f9\u00a00 = n\/2[36 + (\u22122n + 2)]<\/p>\n<p>\u27f9\u00a00 = n\/2[38 \u2212 2n]\u00a0Further,\u00a0n\/2<\/p>\n<p>\u27f9 n = 0 Or, 38 \u2013 2n = 0<\/p>\n<p>\u27f9 2n = 38<\/p>\n<p>\u27f9 n = 19<\/p>\n<p>Since the number of terms cannot be zer0, hence the number of terms (n) should be 19.<\/p>\n<p>(ii) Given, the first term (a) = -14, Filth term (a<sub>5<\/sub>) = 2, Sum of terms (S<sub>n<\/sub>) = 40 of the A.P.<\/p>\n<p>If the common difference is taken as d.<\/p>\n<p>Then, a<sub>5<\/sub>\u00a0= a+ 4d<\/p>\n<p>\u27f9 2 = -14 + 4d<\/p>\n<p>\u27f9 2 + 14 = 4d<\/p>\n<p>\u27f9 4d = 16<\/p>\n<p>\u27f9 d = 4<\/p>\n<p>Next, we know that S<sub>n\u00a0<\/sub>= n\/2[2a + (n \u2212 1)d]<\/p>\n<p>Where; a = first term for the given A.P.<\/p>\n<p>d = common difference of the given A.P.<\/p>\n<p>n = number of terms<\/p>\n<p>Now, on substituting the values in S<sub>n<\/sub><\/p>\n<p>\u27f9\u00a040 = n\/2[2(\u221214) + (n \u2212 1)(4)]<\/p>\n<p>\u27f9\u00a040 = n\/2[\u221228 + (4n \u2212 4)]<\/p>\n<p>\u27f9\u00a040 = n\/2[\u221232 + 4n]<\/p>\n<p>\u27f9 40(2) = \u2013 32n + 4n<sup>2<\/sup><\/p>\n<p>So, we get the following quadratic equation,<\/p>\n<p>4n<sup>2<\/sup>\u00a0\u2013 32n \u2013 80 = 0<\/p>\n<p>\u27f9 n<sup>2<\/sup>\u00a0\u2013 8n \u2013 20 = 0<\/p>\n<p>On solving by factorization method, we get<\/p>\n<p>n<sup>2<\/sup>\u00a0\u2013 10n + 2n \u2013 20 = 0<\/p>\n<p>\u27f9 n(n \u2013 10) + 2(n \u2013 10 ) = 0<\/p>\n<p>\u27f9 (n + 2)(n \u2013 10) = 0<\/p>\n<p>Either, n + 2 = 0<\/p>\n<p>\u27f9 n = -2<\/p>\n<p>Or, n \u2013 10 = 0<\/p>\n<p>\u27f9 n = 10<\/p>\n<p>Since the number of terms cannot be negative<strong>.<\/strong><\/p>\n<p>Therefore, the number of terms (n) is 10.<\/p>\n<p>(iii) Given AP is 9, 17, 25,\u2026<\/p>\n<p>We know that,<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2[2a + (n \u2212 1)d]<\/p>\n<p>Here we have,<\/p>\n<p>The first term (a) = 9 and the sum of n terms (S<sub>n<\/sub>) = 636<\/p>\n<p>Common difference of the A.P. (d) = a<sub>2<\/sub> \u2013 a<sub>1<\/sub>\u00a0= 17 \u2013 9 = 8<\/p>\n<p>Substituting the values in S<sub>n<\/sub>, we get<\/p>\n<p>\u27f9\u00a0636 = n\/2[2(9) + (n \u2212 1)(8)]<\/p>\n<p>\u27f9\u00a0636 = n\/2[18 + (8n \u2212 8)]<\/p>\n<p>\u27f9 636(2) = (n)[10 + 8n]<\/p>\n<p>\u27f9 1271 = 10n + 8n<sup>2<\/sup><\/p>\n<p>Now, we get the following quadratic equation,<\/p>\n<p>\u27f9 8n<sup>2<\/sup>\u00a0+ 10n \u2013 1272 = 0<\/p>\n<p>\u27f9 4n<sup>2<\/sup>+ 5n \u2013 636 = 0<\/p>\n<p>On solving by factorisation method, we have<\/p>\n<p>\u27f9 4n<sup>2<\/sup>\u00a0\u2013 48n + 53n \u2013 636 = 0<\/p>\n<p>\u27f9 4n(n \u2013 12) + 53(n \u2013 12) = 0<\/p>\n<p>\u27f9 (4n + 53)(n \u2013 12) = 0<\/p>\n<p>Either 4n + 53 = 0 \u27f9\u00a0n = -53\/4<\/p>\n<p>Or, n \u2013 12 = 0 \u27f9 n = 12<\/p>\n<p>Since, the number of terms cannot be a fraction.<\/p>\n<p>Therefore, the number of terms (n) is 12.<\/p>\n<p>(iv) Given A.P. is 63, 60, 57,\u2026<\/p>\n<p>We know that,<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2[2a + (n \u2212 1)d]<\/p>\n<p>Here we have,<\/p>\n<p>the first term (a) = 63<\/p>\n<p>The sum of n terms (S<sub>n<\/sub>) = 693<\/p>\n<p>Common difference of the A.P. (d) = a<sub>2<\/sub>\u00a0\u2013 a<sub>1<\/sub>\u00a0= 60 \u2013 63 = \u20133<\/p>\n<p>On substituting the values in S<sub>n\u00a0<\/sub>we get<\/p>\n<p>\u27f9\u00a0693 = n\/2[2(63) + (n \u2212 1)(\u22123)]<\/p>\n<p>\u27f9\u00a0693 = n\/2[126+(\u22123n + 3)]<\/p>\n<p>\u27f9\u00a0693 = n\/2[129 \u2212 3n]<\/p>\n<p>\u27f9 693(2) = 129n \u2013 3n<sup>2<\/sup><\/p>\n<p>Now, we get the following quadratic equation.<\/p>\n<p>\u27f9 3n<sup>2<\/sup>\u00a0\u2013 129n + 1386 = 0<\/p>\n<p>\u27f9 n<sup>2<\/sup>\u00a0\u2013 43n + 462<\/p>\n<p>Solving by factorisation method, we have<\/p>\n<p>\u27f9 n<sup>2<\/sup>\u00a0\u2013 22n \u2013 21n + 462 = 0<\/p>\n<p>\u27f9 n(n \u2013 22) -21(n \u2013 22) = 0<\/p>\n<p>\u27f9 (n \u2013 22) (n \u2013 21) = 0<\/p>\n<p>Either, n \u2013 22 = 0 \u27f9 n = 22<\/p>\n<p>Or, n \u2013 21 = 0 \u27f9 n = 21<\/p>\n<p>Now, the 22<sup>nd<\/sup>\u00a0term will be a<sub>22<\/sub>\u00a0= a<sub>1<\/sub>\u00a0+ 21d = 63 + 21( -3 ) = 63 \u2013 63 = 0<\/p>\n<p>So, the sum of 22 as well as 21 terms is 693.<\/p>\n<p>Therefore, the number of terms (n) is 21 or 22.<\/p>\n<p>(v) Given A.P. is 27, 24, 21. . .<\/p>\n<p>We know that,<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2[2a + (n \u2212 1)d]<\/p>\n<p>Here we have, the first term (a) = 27<\/p>\n<p>The sum of n terms (S<sub>n<\/sub>) = 0<\/p>\n<p>Common difference of the A.P. (d) = a<sub>2<\/sub>\u00a0\u2013 a<sub>1<\/sub>\u00a0= 24 \u2013 27 = -3<\/p>\n<p>On substituting the values in S<sub>n<\/sub>, we get<\/p>\n<p>\u27f9\u00a00 = n\/2[2(27) + (n \u2212 1)( \u2212 3)]<\/p>\n<p>\u27f9 0 = (n)[54 + (n \u2013 1)(-3)]<\/p>\n<p>\u27f9 0 = (n)[54 \u2013 3n + 3]<\/p>\n<p>\u27f9 0 = n [57 \u2013 3n] Further we have, n = 0 Or, 57 \u2013 3n = 0<\/p>\n<p>\u27f9 3n = 57<\/p>\n<p>\u27f9 n = 19<\/p>\n<p>The number of terms cannot be zero,<\/p>\n<p>Hence, the numbers of terms (n) is 19.<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q11. <\/strong><\/p>\n<p><strong>Find the sum of the first<\/strong><\/p>\n<p><strong>(i) 11 terms of the A.P. : 2, 6, 10, 14, . . .\u00a0<\/strong><\/p>\n<p><strong>(ii) 13 terms of the A.P. : -6, 0, 6, 12, . . .\u00a0<\/strong><\/p>\n<p><strong>(iii) 51 terms of the A.P.: whose second term is 2 and the fourth term is 8.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that the sum of terms for different arithmetic progressions is given by<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2[2a + (n \u2212 1)d]<\/p>\n<p>Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms<\/p>\n<p>(i) Given A.P 2, 6, 10, 14,\u2026 to 11 terms.<\/p>\n<p>Common difference (d) = a<sub>2<\/sub>\u00a0\u2013 a<sub>1<\/sub>\u00a0= 10 \u2013 6 = 4<\/p>\n<p>Number of terms (n) = 11<\/p>\n<p>First term for the given A.P. (a) = 2<\/p>\n<p>So,<\/p>\n<p>S<sub>11<\/sub> = 11\/2[2(2) + (11 \u2212 1)4]<\/p>\n<p>=\u00a011\/2[2(2) + (10)4]<\/p>\n<p>=\u00a011\/2[4 + 40]<\/p>\n<p>= 11 \u00d7 22<\/p>\n<p>= 242<\/p>\n<p>Hence, the sum of first 11 terms for the given A.P. is 242<\/p>\n<p>(ii) Given A.P. \u2013 6, 0, 6, 12, \u2026 to 13 terms.<\/p>\n<p>Common difference (d) = a<sub>2<\/sub>\u00a0\u2013 a<sub>1<\/sub>\u00a0= 6 \u2013 0 = 6<\/p>\n<p>Number of terms (n) = 13<\/p>\n<p>First term (a) = -6<\/p>\n<p>So,<\/p>\n<p>S<sub>13<\/sub> = 13\/2[2(\u2212 6) + (13 \u20131)6]<\/p>\n<p>=\u00a013\/2[(\u221212) + (12)6]<\/p>\n<p>=\u00a013\/2[60]\u00a0= 390<\/p>\n<p>Hence, the sum of first 13 terms for the given A.P. is 390<\/p>\n<p>(iii) 51 terms of an AP whose a<sub>2<\/sub>\u00a0= 2 and a<sub>4<\/sub>\u00a0= 8<\/p>\n<p>We know that, a<sub>2<\/sub>\u00a0= a + d<\/p>\n<p>2 = a + d \u2026(2)<\/p>\n<p>Also, a<sub>4<\/sub>\u00a0= a + 3d<\/p>\n<p>8 = a + 3d \u2026 (2)<\/p>\n<p>Subtracting (1) from (2), we have<\/p>\n<p>2d = 6<\/p>\n<p>d = 3<\/p>\n<p>Substituting d = 3 in (1), we get<\/p>\n<p>2 = a + 3<\/p>\n<p>\u27f9 a = -1<\/p>\n<p>Given that the number of terms (n) = 51<\/p>\n<p>First term (a) = -1<\/p>\n<p>So,<\/p>\n<p>S<sub>n<\/sub> = 51\/2[2(\u22121) + (51 \u2212 1)(3)]<\/p>\n<p>=\u00a051\/2[\u22122 + 150]<\/p>\n<p>=\u00a051\/2[148]<\/p>\n<p>= 3774<\/p>\n<p>Hence, the sum of the first 51 terms for the A.P. is 3774.<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q12. <\/strong><\/p>\n<p><strong>Find the sum of<\/strong><\/p>\n<p><strong>(i) the first 15 multiples of 8<\/strong><\/p>\n<p><strong>(ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.<\/strong><\/p>\n<p><strong>(iii) all 3 \u2013 digit natural numbers which are divisible by 13.<\/strong><\/p>\n<p><strong>(iv) all 3 \u2013 digit natural numbers which are multiples of 11.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that the sum of terms for an A.P is given by<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2[2a + (n \u2212 1)d]<\/p>\n<p>Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms<\/p>\n<p>(i) Given, first 15 multiples of 8.<\/p>\n<p>These multiples form an A.P: 8, 16, 24, \u2026\u2026, 120<\/p>\n<p>Here, a = 8 , d = 61 \u2013 8 = 8 and the number of terms(n) = 15<\/p>\n<p>Now, finding the sum of 15 terms, we have<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 9\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-9-18.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 9\" \/>\\<\/p>\n<p>Hence, the sum of the first 15 multiples of 8 is 960<\/p>\n<p>(ii)(a) First 40 positive integers divisible by 3.<\/p>\n<p>Hence, the first multiple is 3 and the 40<sup>th<\/sup>\u00a0multiple is 120.<\/p>\n<p>And, these terms will form an A.P. with the common difference of 3.<\/p>\n<p>Here, First term (a) = 3<\/p>\n<p>Number of terms (n) = 40<\/p>\n<p>Common difference (d) = 3<\/p>\n<p>So, the sum of 40 terms<\/p>\n<p>S<sub>40<\/sub> = 40\/2[2(3) + (40 \u2212 1)3]<\/p>\n<p>= 20[6 + (39)3]<\/p>\n<p>= 20(6 + 117)<\/p>\n<p>= 20(123) = 2460<\/p>\n<p>Thus, the sum of the first 40 multiples of 3 is 2460.<\/p>\n<p>(b) First 40 positive integers divisible by 5<\/p>\n<p>Hence, the first multiple is 5 and the 40<sup>th<\/sup>\u00a0multiple is 200.<\/p>\n<p>And, these terms will form an A.P. with the common difference of 5.<\/p>\n<p>Here, First term (a) = 5<\/p>\n<p>Number of terms (n) = 40<\/p>\n<p>Common difference (d) = 5<\/p>\n<p>So, the sum of 40 terms<\/p>\n<p>S<sub>40 <\/sub>= 40\/2[2(5) + (40 \u2212 1)5]<\/p>\n<p>= 20[10 + (39)5]<\/p>\n<p>= 20 (10 + 195)<\/p>\n<p>= 20 (205) = 4100<\/p>\n<p>Hence, the sum of the first 40 multiples of 5 is 4100.<\/p>\n<p>(c) First 40 positive integers divisible by 6<\/p>\n<p>Hence, the first multiple is 6 and the 40<sup>th<\/sup>\u00a0multiple is 240.<\/p>\n<p>And, these terms will form an A.P. with the common difference of 6.<\/p>\n<p>Here, First term (a) = 6<\/p>\n<p>Number of terms (n) = 40<\/p>\n<p>Common difference (d) = 6<\/p>\n<p>So, the sum of 40 terms<\/p>\n<p>S<sub>40<\/sub> = 40\/2[2(6) + (40 \u2212 1)6]<\/p>\n<p>= 20[12 + (39)6]<\/p>\n<p>=20(12 + 234)<\/p>\n<p>= 20(246) = 4920<\/p>\n<p>Hence, the sum of the first 40 multiples of 6 is 4920.<\/p>\n<p>(iii) All 3 digit natural numbers are divisible by 13.<\/p>\n<p>So, we know that the first 3 digit multiple of 13 is 104 and the last 3 digit multiple of 13 is 988.<\/p>\n<p>And, these terms form an A.P. with the common difference of 13.<\/p>\n<p>Here, first term (a) = 104 and the last term (l) = 988<\/p>\n<p>Common difference (d) = 13<\/p>\n<p>Finding the number of terms in the A.P. by, a<sub>n<\/sub>\u00a0= a + (n \u2212 1)d<\/p>\n<p>We have,<\/p>\n<p>988 = 104 + (n \u2013 1)13<\/p>\n<p>\u27f9 988 = 104 + 13n -13<\/p>\n<p>\u27f9 988 = 91 + 13n<\/p>\n<p>\u27f9 13n = 897<\/p>\n<p>\u27f9 n = 69<\/p>\n<p>Now, using the formula for the sum of n terms, we get<\/p>\n<p>S<sub>69<\/sub> = 69\/2[2(104) + (69 \u2212 1)13]<\/p>\n<p>=\u00a069\/2[208 + 884]<\/p>\n<p>=\u00a069\/2[1092]<\/p>\n<p>= 69(546)<\/p>\n<p>= 37674<\/p>\n<p>Hence, the sum of all 3 digit multiples of 13 is 37674.<\/p>\n<p>(iv) All 3 digit natural numbers which are multiples of 11.<\/p>\n<p>So, we know that the first 3 digit multiple of 11 is 110 and the last 3 digit multiple of 13 is 990.<\/p>\n<p>And, these terms form an A.P. with the common difference of 11.<\/p>\n<p>Here, first term (a) = 110 and the last term (l) = 990<\/p>\n<p>Common difference (d) = 11<\/p>\n<p>Finding the number of terms in the A.P. by, a<sub>n<\/sub>\u00a0= a + (n \u2212 1)d<\/p>\n<p>We get,<\/p>\n<p>990 = 110 + (n \u2013 1)11<\/p>\n<p>\u27f9 990 = 110 + 11n -11<\/p>\n<p>\u27f9 990 = 99 + 11n<\/p>\n<p>\u27f9 11n = 891<\/p>\n<p>\u27f9 n = 81<\/p>\n<p>Now, using the formula for the sum of n terms, we get<\/p>\n<p>S<sub>81<\/sub> = 81\/2[2(110) + (81 \u2212 1)11]<\/p>\n<p>=\u00a081\/2[220 + 880]<\/p>\n<p>=\u00a081\/2[1100]<\/p>\n<p>= 81(550)<\/p>\n<p>= 44550<\/p>\n<p>Hence, the sum of all 3 digit multiples of 11 is 44550.<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q13. <\/strong><\/p>\n<p><strong>Find the sum:<\/strong><\/p>\n<p><strong>(i) 2 + 4 + 6 + . . . + 200\u00a0<\/strong><\/p>\n<p><strong>(ii) 3 + 11 + 19 + . . . + 803\u00a0<\/strong><\/p>\n<p><strong>(iii) (-5) + (-8) + (-11) + . . . + (- 230)\u00a0<\/strong><\/p>\n<p><strong>(iv) 1 + 3 + 5 + 7 + . . . + 199\u00a0<\/strong><\/p>\n<p><strong><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 10\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-9-19.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 10\" \/><\/strong><\/p>\n<p><strong>(vi) 34 + 32 + 30 + . . . + 10\u00a0<\/strong><\/p>\n<p><strong>(vii) 25 + 28 + 31 + . . . + 100\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that the sum of terms for an A.P is given by<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2[2a + (n \u2212 1)d]<\/p>\n<p>Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms<\/p>\n<p>Or S<sub>n<\/sub>\u00a0= n\/2[a + l]<\/p>\n<p>Where; a = first term for the given A.P. ;l = last term for the given A.P<\/p>\n<p>(i) Given series. 2 + 4 + 6 + . . . + 200\u00a0which is an A.P<\/p>\n<p>Where, a = 2 ,d = 4 \u2013 2 = 2 and last term (a<sub>n\u00a0<\/sub>= l) = 200<\/p>\n<p>We know that, a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>So,<\/p>\n<p>200 = 2 + (n \u2013 1)2<\/p>\n<p>200 = 2 + 2n \u2013 2<\/p>\n<p>n = 200\/2 = 100<\/p>\n<p>Now, for the sum of these 100 terms<\/p>\n<p>S<sub>100\u00a0<\/sub>= 100\/2 [2 + 200]<\/p>\n<p>= 50(202)<\/p>\n<p>= 10100<\/p>\n<p>Hence, the sum of terms of the given series is 10100.<\/p>\n<p>(ii) Given series. 3 + 11 + 19 + . . . + 803\u00a0which is an A.P<\/p>\n<p>Where, a = 3 ,d = 11 \u2013 3 = 8 and last term (a<sub>n\u00a0<\/sub>= l) = 803<\/p>\n<p>We know that, a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>So,<\/p>\n<p>803 = 3 + (n \u2013 1)8<\/p>\n<p>803 = 3 + 8n \u2013 8<\/p>\n<p>n = 808\/8 = 101<\/p>\n<p>Now, for the sum of these 101 terms<\/p>\n<p>S<sub>101\u00a0<\/sub>= 101\/2 [3 + 803]<\/p>\n<p>= 101(806)\/2<\/p>\n<p>= 101 x 403<\/p>\n<p>= 40703<\/p>\n<p>Hence, the sum of terms of the given series is 40703.<\/p>\n<p>(iii) Given series (-5) + (-8) + (-11) + . . . + (- 230)which is an A.P<\/p>\n<p>Where, a = -5 ,d = -8 \u2013 (-5) = -3 and last term (a<sub>n\u00a0<\/sub>= l) = -230<\/p>\n<p>We know that, a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>So,<\/p>\n<p>-230 = -5 + (n \u2013 1)(-3)<\/p>\n<p>-230 = -5 \u2013 3n + 3<\/p>\n<p>3n = -2 + 230<\/p>\n<p>n = 228\/3 = 76<\/p>\n<p>Now, for the sum of these 76 terms<\/p>\n<p>S<sub>76\u00a0<\/sub>= 76\/2 [-5 + (-230)]<\/p>\n<p>= 38 x (-235)<\/p>\n<p>= -8930<\/p>\n<p>Hence, the sum of terms of the given series is -8930.<\/p>\n<p>(iv) Given series. 1 + 3 + 5 + 7 + . . . + 199which is an A.P<\/p>\n<p>Where, a = 1 ,d = 3 \u2013 1 = 2 and last term (a<sub>n\u00a0<\/sub>= l) = 199<\/p>\n<p>We know that, a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>So,<\/p>\n<p>199 = 1 + (n \u2013 1)2<\/p>\n<p>199 = 1 + 2n \u2013 2<\/p>\n<p>n = 200\/2 = 100<\/p>\n<p>Now, for the sum of these 100 terms<\/p>\n<p>S<sub>100\u00a0<\/sub>= 100\/2 [1 + 199]<\/p>\n<p>= 50(200)<\/p>\n<p>= 10000<\/p>\n<p>Hence, the sum of terms of the given series is 10000.<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 11\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-9-20.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 11\" \/><\/p>\n<p>(v) Given series which is an A.P<\/p>\n<p>Where, a = 7, d = 10 \u00bd \u2013 7 = (21 \u2013 14)\/2 = 7\/2 and last term (a<sub>n\u00a0<\/sub>= l) = 84<\/p>\n<p>We know that, a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>So,<\/p>\n<p>84 = 7 + (n \u2013 1)(7\/2)<\/p>\n<p>168 = 14 + 7n \u2013 7<\/p>\n<p>n = (168 \u2013 7)\/7 = 161\/7 = 23<\/p>\n<p>Now, for the sum of these 23 terms<\/p>\n<p>S<sub>23\u00a0<\/sub>= 23\/2 [7 + 84]<\/p>\n<p>= 23(91)\/2<\/p>\n<p>= 2093\/2<\/p>\n<p>Hence, the sum of terms of the given series is 2093\/2.<\/p>\n<p>(vi) Given series, 34 + 32 + 30 + . . . + 10which is an A.P<\/p>\n<p>Where, a = 34 ,d = 32 \u2013 34 = -2 and last term (a<sub>n\u00a0<\/sub>= l) = 10<\/p>\n<p>We know that, a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>So,<\/p>\n<p>10 = 34 + (n \u2013 1)(-2)<\/p>\n<p>10 = 34 \u2013 2n + 2<\/p>\n<p>n = (36 \u2013 10)\/2 = 13<\/p>\n<p>Now, for the sum of these 13 terms<\/p>\n<p>S<sub>13\u00a0<\/sub>= 13\/2 [34 + 10]<\/p>\n<p>= 13(44)\/2<\/p>\n<p>= 13 x 22<\/p>\n<p>= 286<\/p>\n<p>Hence, the sum of terms of the given series is 286.<\/p>\n<p>(vii) Given series, 25 + 28 + 31 + . . . + 100which is an A.P<\/p>\n<p>Where, a = 25 ,d = 28 \u2013 25 = 3 and last term (a<sub>n\u00a0<\/sub>= l) = 100<\/p>\n<p>We know that, a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>So,<\/p>\n<p>100 = 25 + (n \u2013 1)(3)<\/p>\n<p>100 = 25 + 3n \u2013 3<\/p>\n<p>n = (100 \u2013 22)\/3 = 26<\/p>\n<p>Now, for the sum of these 26 terms<\/p>\n<p>S<sub>100\u00a0<\/sub>= 26\/2 [25 + 100]<\/p>\n<p>= 13(125)<\/p>\n<p>= 1625<\/p>\n<p>Hence, the sum of terms of the given series is 1625.<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q14. <\/strong><\/p>\n<p><strong>The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given, the first term of the A.P (a) = 17<\/p>\n<p>The last term of the A.P (l) = 350<\/p>\n<p>The common difference (d) of the A.P. = 9<\/p>\n<p>Let the number of terms be n. And, we know that; l = a + (n \u2013 1)d<\/p>\n<p>So, 350 = 17 + (n- 1) 9<\/p>\n<p>\u27f9 350 = 17 + 9n \u2013 9<\/p>\n<p>\u27f9 350 = 8 + 9n<\/p>\n<p>\u27f9 350 \u2013 8 = 9n<\/p>\n<p>Thus we get, n = 38<\/p>\n<p>Now, finding the sum of terms<\/p>\n<p>S<sub>n<\/sub> = n\/2[a + l]<\/p>\n<p>= 38\/2(17 + 350)<\/p>\n<p>= 19 \u00d7 367<\/p>\n<p>= 6973<\/p>\n<p>Hence, the number of terms is of the A.P is 38 and their sum is 6973.<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q15. <\/strong><\/p>\n<p><strong>The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference, and the sum of the first 20 terms.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s consider the first term as a and the common difference as d.<\/p>\n<p>Given,<\/p>\n<p>a<sub>3<\/sub>\u00a0= 7\u00a0\u2026. (1) and,<\/p>\n<p>a<sub>7<\/sub>\u00a0= 3a<sub>3<\/sub>\u00a0+ 2 \u2026. (2)<\/p>\n<p>So, using (1) in (2), we get,<\/p>\n<p>a<sub>7\u00a0<\/sub>= 3(7) + 2 = 21 + 2 = 23 \u2026. (3)<\/p>\n<p>Also, we know that<\/p>\n<p>a<sub>n<\/sub>\u00a0= a +(n \u2013 1)d<\/p>\n<p>So, the 3th term (for n = 3),<\/p>\n<p>a<sub>3<\/sub>\u00a0= a + (3 \u2013 1)d<\/p>\n<p>\u27f9 7 = a + 2d (Using 1)<\/p>\n<p>\u27f9 a = 7 \u2013 2d \u2026. (4)<\/p>\n<p>Similarly, for the 7th term (n = 7),<\/p>\n<p>a<sub>7<\/sub>\u00a0= a + (7 \u2013 1) d 24 = a + 6d\u00a0= 23 (Using 3)<\/p>\n<p>a = 23 \u2013 6d \u2026. (5)<\/p>\n<p>Subtracting (4) from (5), we get,<\/p>\n<p>a \u2013 a = (23 \u2013 6d) \u2013 (7 \u2013 2d)<\/p>\n<p>\u27f9 0 = 23 \u2013 6d \u2013 7 + 2d<\/p>\n<p>\u27f9 0 = 16 \u2013 4d<\/p>\n<p>\u27f9 4d = 16<\/p>\n<p>\u27f9 d = 4<\/p>\n<p>Now, to find a, we substitute the value of d in (4), a =7 \u2013 2(4)<\/p>\n<p>\u27f9 a = 7 \u2013 8<\/p>\n<p>a = -1<\/p>\n<p>Hence, for the A.P. a = -1 and d = 4<\/p>\n<p>For finding the sum, we know that<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2[2a + (n \u2212 1)d]\u00a0and n = 20 (given)<\/p>\n<p>S<sub>20<\/sub> = 20\/2[2(\u22121) + (20 \u2212 1)(4)]<\/p>\n<p>= (10)[-2 + (19)(4)]<\/p>\n<p>= (10)[-2 + 76]<\/p>\n<p>= (10)[74]<\/p>\n<p>= 740<\/p>\n<p>Hence, the sum of the first 20 terms for the given A.P. is 740<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q16. <\/strong><\/p>\n<p><strong>The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The first term of the A.P (a) = 2<\/p>\n<p>The last term of the A.P (l) = 50<\/p>\n<p>Sum of all the terms S<sub>n<\/sub>\u00a0= 442<\/p>\n<p>So, let the common difference of the A.P. be taken as d.<\/p>\n<p>The sum of all the terms is given as,<\/p>\n<p>442 = (n\/2)(2 + 50)<\/p>\n<p>\u27f9\u00a0442 = (n\/2)(52)<\/p>\n<p>\u27f9 26n = 442<\/p>\n<p>\u27f9 n = 17<\/p>\n<p>Now, the last term is expressed as<\/p>\n<p>50 = 2 + (17 \u2013 1)d<\/p>\n<p>\u27f9 50 = 2 + 16d<\/p>\n<p>\u27f9 16d = 48<\/p>\n<p>\u27f9 d = 3<\/p>\n<p>Thus, the common difference of the A.P. is d = 3.<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q17. <\/strong><\/p>\n<p><strong>If the 12<sup>th<\/sup> term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of the first 10 terms?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us take the first term as a and the common difference as d.<\/p>\n<p>Given,<\/p>\n<p>a<sub>12<\/sub>\u00a0= -13 S<sub>4<\/sub>\u00a0= 24<\/p>\n<p>Also, we know that a<sub>n\u00a0<\/sub>= a + (n \u2013 1)d<\/p>\n<p>So, for the 12th term<\/p>\n<p>a<sub>12<\/sub>\u00a0= a + (12 \u2013 1)d = -13<\/p>\n<p>\u27f9 a + 11d = -13<\/p>\n<p>a = -13 \u2013 11d \u2026. (1)<\/p>\n<p>And, we that for sum of terms<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2[2a + (n \u2212 1)d]<\/p>\n<p>Here, n = 4<\/p>\n<p>S<sub>4<\/sub> = 4\/2[2(a) + (4 \u2212 1)d]<\/p>\n<p>\u27f9 24 = (2)[2a + (3)(d)]<\/p>\n<p>\u27f9 24 = 4a + 6d<\/p>\n<p>\u27f9 4a = 24 \u2013 6d<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 12\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/https-files-askiitians-com-cdn-images-2018917-14.png\" alt=\"https:\/\/files.askiitians.com\/cdn\/images\/2018917-14419946-576-equation-10.png\" \/><\/p>\n<p>Subtracting (1) from (2), we have<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 12\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-9-21.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 12\" \/><\/p>\n<p>Further simplifying for d, we get,<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-9-22.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 13\" \/><\/p>\n<p>\u27f9 -19 \u00d7 2 = 19d<\/p>\n<p>\u27f9 d = \u2013 2<\/p>\n<p>On substituting the value of d in (1), we find a<\/p>\n<p>a = -13 \u2013 11(-2)<\/p>\n<p>a = -13 + 22<\/p>\n<p>a = 9<\/p>\n<p>Next, the sum of 10 term is given by<\/p>\n<p>S<sub>10<\/sub> = 10\/2[2(9) + (10 \u2212 1)(\u22122)]<\/p>\n<p>= (5)[19 + (9)(-2)]<\/p>\n<p>= (5)(18 \u2013 18) = 0<\/p>\n<p>Thus, the sum of the first 10 terms for the given A.P. is S<sub>10<\/sub>\u00a0= 0.<\/p>\n<p><strong>18.<\/strong><\/p>\n<p><img class=\"\" title=\"Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - q18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/class-10-maths-chapter-9-arithmetic-ex-6-q18-1.png\" alt=\"Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - q18\" width=\"309\" height=\"244\" \/><\/p>\n<p><img class=\"\" title=\"Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - q18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/class-10-maths-chapter-9-arithmetic-ex-6-q18-2.png\" alt=\"Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - q18\" width=\"352\" height=\"290\" \/><\/p>\n<p><strong>19. In an A.P., if the first term is 22, the common difference is \u2013 4, and the sum to n terms is 64, find n.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given that,<\/p>\n<p>a = 22, d = \u2013 4 and S<sub>n<\/sub>\u00a0= 64<\/p>\n<p>Let us consider the number of terms as n.<\/p>\n<p>For sum of terms in an A.P, we know that<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2[2a + (n \u2212 1)d]<\/p>\n<p>Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms<\/p>\n<p>So,<\/p>\n<p>\u27f9\u00a0S<sub>n<\/sub>\u00a0= n\/2[2(22) + (n \u2212 1)(\u22124)]<\/p>\n<p>\u27f9\u00a064 = n\/2[2(22) + (n \u2212 1)(\u22124)]<\/p>\n<p>\u27f9 64(2) = n(48 \u2013 4n)<\/p>\n<p>\u27f9 128 = 48n \u2013 4n<sup>2<\/sup><\/p>\n<p>After rearranging the terms, we have a quadratic equation<\/p>\n<p>4n<sup>2<\/sup>\u00a0\u2013 48n + 128 = 0,<\/p>\n<p>n<sup>2<\/sup>\u00a0\u2013 12n + 32 = 0 [dividing by 4 on both sides]<\/p>\n<p>n<sup>2<\/sup>\u00a0\u2013 12n + 32 = 0<\/p>\n<p>Solving by factorisation method,<\/p>\n<p>n<sup>2<\/sup>\u00a0\u2013 8n \u2013 4n + 32 = 0<\/p>\n<p>n ( n \u2013 8 ) \u2013 4 ( n \u2013 8 ) = 0<\/p>\n<p>(n \u2013 8) (n \u2013 4) = 0<\/p>\n<p>So, we get n \u2013 8 = 0 \u27f9 n = 8<\/p>\n<p>Or, n \u2013 4 = 0 \u27f9 n = 4<\/p>\n<p>Hence, the number of terms can be either n = 4 or 8.<\/p>\n<p><strong>20. In an A.P., if the 5<sup>th<\/sup>\u00a0and 12<sup>th<\/sup>\u00a0terms are 30 and 65 respectively, what is the sum of first 20 terms?\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s take the first term as a and the common difference to be d<\/p>\n<p>Given that,<\/p>\n<p>a<sub>5<\/sub>\u00a0= 30 and a<sub>12<\/sub>\u00a0= 65<\/p>\n<p>And, we know that a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>So,<\/p>\n<p>a<sub>5<\/sub>\u00a0= a + (5 \u2013 1)d<\/p>\n<p>30 = a + 4d<\/p>\n<p>a = 30 \u2013 4d \u2026. (i)<\/p>\n<p>Similarly, a<sub>12<\/sub>\u00a0= a + (12 \u2013 1) d<\/p>\n<p>65 = a + 11d<\/p>\n<p>a = 65 \u2013 11d \u2026. (ii)<\/p>\n<p>Subtracting (i) from (ii), we have<\/p>\n<p>a \u2013 a = (65 \u2013 11d) \u2013 (30 \u2013 4d)<\/p>\n<p>0 = 65 \u2013 11d \u2013 30 + 4d<\/p>\n<p>0 = 35 \u2013 7d<\/p>\n<p>7d = 35<\/p>\n<p>d = 5<\/p>\n<p>Putting d in (i), we get<\/p>\n<p>a = 30 \u2013 4(5)<\/p>\n<p>a = 30 \u2013 20<\/p>\n<p>a = 10<\/p>\n<p>Thus for the A.P; d = 5 and a = 10<\/p>\n<p>Next, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2[2a + (n \u2212 1)d]<\/p>\n<p>Where;<\/p>\n<p>a = first term of the given A.P.<\/p>\n<p>d = common difference of the given A.P.<\/p>\n<p>n = number of terms<\/p>\n<p>Here n = 20, so we have<\/p>\n<p>S<sub>20\u00a0<\/sub>= 20\/2[2(10) + (20 \u2212 1)(5)]<\/p>\n<p>= (10)[20 + (19)(5)]<\/p>\n<p>= (10)[20 + 95]<\/p>\n<p>= (10)[115]<\/p>\n<p>= 1150<\/p>\n<p>Hence, the sum of the first 20 terms for the given A.P. is 1150<\/p>\n<p><strong>21. Find the sum of the first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s take the first term as a and the common difference as d.<\/p>\n<p>Given that,<\/p>\n<p>a<sub>2<\/sub>\u00a0= 14 and a<sub>3<\/sub>\u00a0= 18<\/p>\n<p>And, we know that a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>So,<\/p>\n<p>a<sub>2<\/sub>\u00a0= a + (2 \u2013 1)d<\/p>\n<p>\u27f9 14 = a + d<\/p>\n<p>\u27f9 a = 14 \u2013 d \u2026. (i)<\/p>\n<p>Similarly,<\/p>\n<p>a<sub>3<\/sub>\u00a0= a + (3 \u2013 1)d<\/p>\n<p>\u27f9 18 = a + 2d<\/p>\n<p>\u27f9 a = 18 \u2013 2d \u2026. (ii)<\/p>\n<p>Subtracting (i) from (ii), we have<\/p>\n<p>a \u2013 a = (18 \u2013 2d) \u2013 (14 \u2013 d)<\/p>\n<p>0 = 18 \u2013 2d \u2013 14 + d<\/p>\n<p>0 = 4 \u2013 d<\/p>\n<p>d = 4<\/p>\n<p>Putting d in (i), to find a<\/p>\n<p>a = 14 \u2013 4<\/p>\n<p>a = 10<\/p>\n<p>Thus, for the A.P. d = 4 and a = 10<\/p>\n<p>Now, to find sum of terms<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2(2a + (n \u2212 1)d)<\/p>\n<p>Where,<\/p>\n<p>a = the first term of the A.P.<\/p>\n<p>d = common difference of the A.P.<\/p>\n<p>n = number of terms So, using the formula for<\/p>\n<p>n = 51,<\/p>\n<p>\u27f9 S<sub>51<\/sub> = 51\/2[2(10) + (51 \u2013 1)(4)]<\/p>\n<p>=\u00a051\/2[20 + (40)4]<\/p>\n<p>=\u00a051\/2[220]<\/p>\n<p>= 51(110)<\/p>\n<p>= 5610<\/p>\n<p>Hence, the sum of the first 51 terms of the given A.P. is 5610<\/p>\n<p><strong>22. If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The Sum of 7 terms of an A.P. is 49<\/p>\n<p>\u27f9 S<sub>7<\/sub>\u00a0= 49<\/p>\n<p>And, the sum of 17 terms of an A.P. is 289<\/p>\n<p>\u27f9 S<sub>17<\/sub>\u00a0= 289<\/p>\n<p>Let the first term of the A.P be a and common difference as d.<\/p>\n<p>And, we know that the sum of n terms of an A.P is<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2[2a + (n \u2212 1)d]<\/p>\n<p>So,<\/p>\n<p>S<sub>7\u00a0<\/sub>= 49 = 7\/2[2a + (7 \u2013 1)d]<\/p>\n<p>= 7\/2 [2a + 6d]<\/p>\n<p>= 7[a + 3d]<\/p>\n<p>\u27f9 7a + 21d = 49<\/p>\n<p>a + 3d = 7 \u2026.. (i)<\/p>\n<p>Similarly,<\/p>\n<p>S<sub>17\u00a0<\/sub>= 17\/2[2a + (17 \u2013 1)d]<\/p>\n<p>= 17\/2 [2a + 16d]<\/p>\n<p>= 17[a + 8d]<\/p>\n<p>\u27f9 17[a + 8d] = 289<\/p>\n<p>a + 8d = 17 \u2026.. (ii)<\/p>\n<p>Now, subtracting (i) from (ii), we have<\/p>\n<p>a + 8d \u2013 (a + 3d) = 17 \u2013 7<\/p>\n<p>5d = 10<\/p>\n<p>d = 2<\/p>\n<p>Putting d in (i), we find a<\/p>\n<p>a + 3(2) = 7<\/p>\n<p>a = 7 \u2013 6 = 1<\/p>\n<p>So, for the A.P: a = 1 and d = 2<\/p>\n<p>For the sum of n terms is given by,<\/p>\n<p>S<sub>n<\/sub> = n\/2[2(1) + (n \u2212 1)(2)]<\/p>\n<p>= n\/2[2 + 2n \u2013 2]<\/p>\n<p>= n\/2[2n]<\/p>\n<p>= n<sup>2<\/sup><\/p>\n<p>Therefore, the sum of n terms of the A.P is given by n<sup>2<\/sup>.<\/p>\n<p><strong>23. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The Sum of the first n terms of an A.P is given by S<sub>n<\/sub>\u00a0= n\/2(2a + (n \u2212 1)d)<\/p>\n<p>Given,<\/p>\n<p>First term (a) = 5, last term (a<sub>n<\/sub>) = 45 and sum of n terms (S<sub>n<\/sub>) = 400<\/p>\n<p>Now, we know that<\/p>\n<p>a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>\u27f9 45 = 5 + (n \u2013 1)d<\/p>\n<p>\u27f9 40 = nd \u2013 d<\/p>\n<p>\u27f9 nd \u2013 d = 40 \u2026. (1)<\/p>\n<p>Also,<\/p>\n<p>S<sub>n\u00a0<\/sub>= n\/2(2(a) + (n \u2212 1)d)<\/p>\n<p>400 = n\/2(2(5) + (n \u2212 1)d)<\/p>\n<p>800 = n (10 + nd \u2013 d)<\/p>\n<p>800 = n (10 + 40) [using (1)]<\/p>\n<p>\u27f9 n\u00a0= 16<\/p>\n<p>Putting n in (1), we find d<\/p>\n<p>nd \u2013 d = 40<\/p>\n<p>16d \u2013 d = 40<\/p>\n<p>15d = 40<\/p>\n<p>d =\u00a08\/3<\/p>\n<p>Therefore, the common difference of the given A.P. is 8\/3.<\/p>\n<p><strong>24. In an A.P. the first term is 8, the n<sup>th<\/sup> term is 33 and the sum of the first n term is 123. Find n and the d, the common difference.\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The first term of the A.P (a) = 8<\/p>\n<p>The nth term of the A.P (l) = 33<\/p>\n<p>And, the sum of all the terms S<sub>n<\/sub>\u00a0= 123<\/p>\n<p>Let the common difference of the A.P. be d.<\/p>\n<p>So, find the number of terms by<\/p>\n<p>123 = (n\/2)(8 + 33)<\/p>\n<p>123 = (n\/2)(41)<\/p>\n<p>n = (123 x 2)\/ 41<\/p>\n<p>n = 246\/41<\/p>\n<p>n = 6<\/p>\n<p>Next, to find the common difference of the A.P. we know that<\/p>\n<p>l = a + (n \u2013 1)d<\/p>\n<p>33 = 8 + (6 \u2013 1)d<\/p>\n<p>33 = 8 + 5d<\/p>\n<p>5d = 25<\/p>\n<p>d = 5<\/p>\n<p>Thus, the number of terms is n = 6 and the common difference of the A.P. is d = 5.<\/p>\n<p><strong>25. In an A.P. the first term is 22, the n<sup>th<\/sup> term is -11 and the sum of the first n term is 66. Find n and the d, the common difference.\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The first term of the A.P (a) = 22<\/p>\n<p>The nth term of the A.P (l) = -11<\/p>\n<p>And, sum of all the terms S<sub>n<\/sub>\u00a0= 66<\/p>\n<p>Let the common difference of the A.P. be d.<\/p>\n<p>So, finding the number of terms by<\/p>\n<p>66 = (n\/2)[22 + (\u221211)]<\/p>\n<p>66 = (n\/2)[22 \u2212 11]<\/p>\n<p>(66)(2) = n(11)<\/p>\n<p>6 \u00d7 2 = n<\/p>\n<p>n = 12<\/p>\n<p>Now, for finding d<\/p>\n<p>We know that, l = a + (n \u2013 1)d<\/p>\n<p>\u2013 11 = 22 + (12 \u2013 1)d<\/p>\n<p>-11 = 22 + 11d<\/p>\n<p>11d = \u2013 33<\/p>\n<p>d = \u2013 3<\/p>\n<p>Hence, the number of terms is n = 12, and the common difference d = -3<\/p>\n<p><strong>26. The first and the last terms of an A.P. are 7 and 49 respectively. If the sum of all its terms is 420, find the common difference.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>First term (a) = 7, last term (a<sub>n<\/sub>) = 49 and sum of n terms (S<sub>n<\/sub>) = 420<\/p>\n<p>Now, we know that<\/p>\n<p>a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>\u27f9 49 = 7 + (n \u2013 1)d<\/p>\n<p>\u27f9 43 = nd \u2013 d<\/p>\n<p>\u27f9 nd \u2013 d = 42 \u2026.. (1)<\/p>\n<p>Next,<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2(2(7) + (n \u2212 1)d)<\/p>\n<p>\u27f9 840 = n[14 + nd \u2013 d]<\/p>\n<p>\u27f9 840 = n[14 + 42] [using (1)]<\/p>\n<p>\u27f9 840 = 54n<\/p>\n<p>\u27f9 n = 15\u00a0\u2026. (2)<\/p>\n<p>So, by substituting (2) in (1), we have<\/p>\n<p>nd \u2013 d = 42<\/p>\n<p>\u27f9 15d \u2013 d = 42<\/p>\n<p>\u27f9 14d = 42<\/p>\n<p>\u27f9 d = 3<\/p>\n<p>Therefore, the common difference of the given A.P. is 3.<\/p>\n<p><strong>27. The first and the last terms of an A.P are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>First term (a) = 5 and the last term (l) = 45<\/p>\n<p>Also, S<sub>n<\/sub>\u00a0= 400<\/p>\n<p>We know that,<\/p>\n<p>a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>\u27f9 45 = 5 + (n \u2013 1)d<\/p>\n<p>\u27f9 40 = nd \u2013 d<\/p>\n<p>\u27f9 nd \u2013 d = 40 \u2026.. (1)<\/p>\n<p>Next,<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2(2(5) + (n \u2212 1)d)<\/p>\n<p>\u27f9 400 = n[10 + nd \u2013 d]<\/p>\n<p>\u27f9 800 = n[10 + 40] [using (1)]<\/p>\n<p>\u27f9 800 = 50n<\/p>\n<p>\u27f9 n = 16\u00a0\u2026. (2)<\/p>\n<p>So, by substituting (2) in (1), we have<\/p>\n<p>nd \u2013 d = 40<\/p>\n<p>\u27f9 16d \u2013 d = 40<\/p>\n<p>\u27f9 15d = 40<\/p>\n<p>\u27f9 d = 8\/3<\/p>\n<p>Therefore, the common difference of the given A.P. is 8\/3.<\/p>\n<p><strong>28. The sum of the first q terms of an A.P. is 162. The ratio of its 6<sup>th<\/sup>\u00a0term to its 13<sup>th<\/sup>\u00a0term is 1: 2. Find the first and 15<sup>th<\/sup> terms of the A.P.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let a be the first term and d be a common difference.<\/p>\n<p>And we know that, sum of first n terms is:<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2(2a + (n \u2212 1)d)<\/p>\n<p>Also, nth term is given by a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>From the question, we have<\/p>\n<p>S<sub>q<\/sub>\u00a0= 162 and\u00a0a<sub>6\u00a0<\/sub>: a<sub>13<\/sub>\u00a0= 1 : 2<\/p>\n<p>So,<\/p>\n<p>2a<sub>6\u00a0<\/sub>= a<sub>13<\/sub><\/p>\n<p>\u27f9 2 [a + (6 \u2013 1d)] = a + (13 \u2013 1)d<\/p>\n<p>\u27f9 2a + 10d = a + 12d<\/p>\n<p>\u27f9 a = 2d \u2026. (1)<\/p>\n<p>And, S<sub>9<\/sub>\u00a0= 162<\/p>\n<p>\u27f9\u00a0S<sub>9<\/sub>\u00a0= 9\/2(2a + (9 \u2212 1)d)<\/p>\n<p>\u27f9 162 =\u00a09\/2(2a + 8d)<\/p>\n<p>\u27f9 162 \u00d7 2 = 9[4d + 8d] [from (1)]<\/p>\n<p>\u27f9 324 = 9 \u00d7 12d<\/p>\n<p>\u27f9 d = 3<\/p>\n<p>\u27f9 a = 2(3) [from (1)]<\/p>\n<p>\u27f9 a = 6<\/p>\n<p>Hence, the first term of the A.P. is 6<\/p>\n<p>For the 15<sup>th<\/sup>\u00a0term, a<sub>15<\/sub>\u00a0= a + 14d = 6 + 14 \u00d7 3 = 6 + 42<\/p>\n<p>Therefore, a<sub>15<\/sub>\u00a0= 48<\/p>\n<p><strong>29. If the 10<sup>th<\/sup>\u00a0term of an A.P. is 21 and the sum of its first 10 terms is 120, find its n<sup>th<\/sup>\u00a0term.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s consider a to be the first term and d be a common difference.<\/p>\n<p>And we know that, sum of first n terms is:<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2(2a + (n \u2212 1)d)\u00a0and n<sup>th<\/sup>\u00a0term is given by: a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>Now, from the question we have<\/p>\n<p>S<sub>10<\/sub>\u00a0= 120<\/p>\n<p>\u27f9\u00a0120 = 10\/2(2a + (10 \u2212 1)d)<\/p>\n<p>\u27f9\u00a0120 = 5(2a + 9d)<\/p>\n<p>\u27f9 24 = 2a + 9d\u00a0\u2026. (1)<\/p>\n<p>Also given that, a<sub>10<\/sub>\u00a0= 21<\/p>\n<p>\u27f9\u00a021 = a + (10 \u2013 1)d<\/p>\n<p>\u27f9\u00a021 = a + 9d\u00a0\u2026. (2)<\/p>\n<p>Subtracting (2) from (1), we get<\/p>\n<p>24 \u2013 21 = 2a + 9d \u2013 a \u2013 9d<\/p>\n<p>\u27f9a = 3<\/p>\n<p>Now, on putting a = 3 in equation (2), we get<\/p>\n<p>3 + 9d = 21<\/p>\n<p>9d = 18<\/p>\n<p>d = 2<\/p>\n<p>Thus, we have the first term(a) = 3 and the common difference(d) = 2<\/p>\n<p>Therefore, the n<sup>th<\/sup>\u00a0term is given by<\/p>\n<p>a<sub>n<\/sub> = a + (n \u2013 1)d = 3 + (n \u2013 1)2<\/p>\n<p>= 3 + 2n -2<\/p>\n<p>= 2n + 1<\/p>\n<p>Hence, the n<sup>th<\/sup>\u00a0term of the A.P is (a<sub>n<\/sub>) = 2n + 1.<\/p>\n<p><strong>30. The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28<sup>th<\/sup>\u00a0term of this A.P.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s take a to be the first term and d to be a common difference.<\/p>\n<p>And we know that, the sum of first n terms<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2(2a + (n \u2212 1)d)<\/p>\n<p>Given that sum of the first 7 terms of an A.P. is 63.<\/p>\n<p>S<sub>7<\/sub>\u00a0= 63<\/p>\n<p>And sum of next 7 terms is 161.<\/p>\n<p>So, the sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms<\/p>\n<p>S<sub>14\u00a0<\/sub>= 63 + 161 = 224<\/p>\n<p>Now, having<\/p>\n<p>S<sub>7<\/sub>\u00a0= 7\/2(2a + (7 \u2212 1)d)<\/p>\n<p>\u27f9 63(2) = 7(2a + 6d)<\/p>\n<p>\u27f9 9 \u00d7 2 = 2a + 6d<\/p>\n<p>\u27f9 2a + 6d = 18\u00a0. . . . (1)<\/p>\n<p>And,<\/p>\n<p>S<sub>14<\/sub>\u00a0= 14\/2(2a + (14 \u2212 1)d)<\/p>\n<p>\u27f9 224 = 7(2a + 13d)<\/p>\n<p>\u27f9 32 = 2a + 13d\u00a0\u2026. (2)<\/p>\n<p>Now, subtracting (1) from (2), we get<\/p>\n<p>\u27f9 13d \u2013 6d = 32 \u2013 18<\/p>\n<p>\u27f9 7d = 14<\/p>\n<p>\u27f9 d = 2<\/p>\n<p>Using d in (1), we have<\/p>\n<p>2a + 6(2) = 18<\/p>\n<p>2a = 18 \u2013 12<\/p>\n<p>a = 3<\/p>\n<p>Thus, from n<sup>th<\/sup>\u00a0term<\/p>\n<p>\u27f9 a<sub>28 <\/sub>= a + (28 \u2013 1)d<\/p>\n<p>= 3 + 27 (2)<\/p>\n<p>= 3 + 54 = 57<\/p>\n<p>Therefore, the 28<sup>th<\/sup>\u00a0term is 57.<\/p>\n<p><strong>31. The sum of the first seven terms of an A.P. is 182. If its 4<sup>th<\/sup>\u00a0and 17<sup>th<\/sup>\u00a0terms are in ratio 1: 5, find the A.P.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given that,<\/p>\n<p>S<sub>17<\/sub>\u00a0= 182<\/p>\n<p>And, we know that the sum of first n term is:<\/p>\n<p>S<sub>n\u00a0<\/sub>= n\/2(2a + (n \u2212 1)d)<\/p>\n<p>So,<\/p>\n<p>S<sub>7<\/sub>\u00a0= 7\/2(2a + (7 \u2212 1)d)<\/p>\n<p>182 \u00d7 2 = 7(2a + 6d)<\/p>\n<p>364 = 14a + 42d<\/p>\n<p>26 = a + 3d<\/p>\n<p>a = 26 \u2013 3d\u00a0\u2026 (1)<\/p>\n<p>Also, it\u2019s given that 4<sup>th<\/sup>\u00a0term and 17<sup>th<\/sup>\u00a0term are in a ratio of 1: 5. So, we have<\/p>\n<p>\u27f9 5(a<sub>4<\/sub>) = 1(a<sub>17<\/sub>)<\/p>\n<p>\u27f9 5 (a + 3d) = 1 (a + 16d)<\/p>\n<p>\u27f9 5a + 15d = a + 16d<\/p>\n<p>\u27f9 4a = d\u00a0\u2026. (2)<\/p>\n<p>Now, substituting (2) in (1), we get<\/p>\n<p>\u27f9 4 ( 26 \u2013 3d ) = d<\/p>\n<p>\u27f9 104 \u2013 12d = d<\/p>\n<p>\u27f9 104 = 13d<\/p>\n<p>\u27f9 d = 8<\/p>\n<p>Putting d in (2), we get<\/p>\n<p>\u27f9 4a = d<\/p>\n<p>\u27f9 4a = 8<\/p>\n<p>\u27f9 a = 2<\/p>\n<p>Therefore, the first term is 2 and the common difference is 8. So, the A.P. is 2, 10, 18, 26, . ..<\/p>\n<p><strong>32. The n<sup>th<\/sup> term of an A.P is given by (-4n + 15). Find the sum of the first 20 terms of this A.P.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The n<sup>th<\/sup>\u00a0term of the A.P = (-4n + 15)<\/p>\n<p>So, by putting n = 1 and n = 20 we can find the first ans 20<sup>th<\/sup>\u00a0term of the A.P<\/p>\n<p>a = (-4(1) + 15) = 11<\/p>\n<p>And,<\/p>\n<p>a<sub>20<\/sub>\u00a0= (-4(20) + 15) = -65<\/p>\n<p>Now, for find the sum of 20 terms of this A.P we have the first and last term.<\/p>\n<p>So, using the formula<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2(a + l)<\/p>\n<p>S<sub>20<\/sub> = 20\/2(11 + (-65))<\/p>\n<p>= 10(-54)<\/p>\n<p>= -540<\/p>\n<p>Therefore, the sum of the first 20 terms of this A.P. is -540.<\/p>\n<p><strong>33. In an A.P. the sum of the first ten terms is -150 and the sum of its next 10 terms is -550. Find the A.P.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s take a to be the first term and d to be a common difference.<\/p>\n<p>And we know that, the sum of first n terms<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2(2a + (n \u2212 1)d)<\/p>\n<p>Given that sum of the first 10 terms of an A.P. is -150.<\/p>\n<p>S<sub>10<\/sub>\u00a0= -150<\/p>\n<p>And the sum of next 10 terms is -550.<\/p>\n<p>So, the sum of first 20 terms = Sum of first 10 terms + sum of next 10 terms<\/p>\n<p>S<sub>20\u00a0<\/sub>= -150 + -550 = -700<\/p>\n<p>Now, having<\/p>\n<p>S<sub>10<\/sub>\u00a0= 10\/2(2a + (10 \u2212 1)d)<\/p>\n<p>\u27f9 -150 = 5(2a + 9d)<\/p>\n<p>\u27f9 -30 = 2a + 9d<\/p>\n<p>\u27f9 2a + 9d = -30\u00a0. . . . (1)<\/p>\n<p>And,<\/p>\n<p>S<sub>20<\/sub>\u00a0= 20\/2(2a + (20 \u2212 1)d)<\/p>\n<p>\u27f9 -700 = 10(2a + 19d)<\/p>\n<p>\u27f9 -70 = 2a + 19d\u00a0\u2026. (2)<\/p>\n<p>Now, subtracting (1) from (2), we get<\/p>\n<p>\u27f9 19d \u2013 9d = -70 \u2013 (-30)<\/p>\n<p>\u27f9 10d = -40<\/p>\n<p>\u27f9 d = -4<\/p>\n<p>Using d in (1), we have<\/p>\n<p>2a + 9(-4) = -30<\/p>\n<p>2a = -30 + 36<\/p>\n<p>a = 6\/2 = 3<\/p>\n<p>Hence, we have a = 3 and d = -4<\/p>\n<p>So, the A.P is 3, -1, -5, -9, -13,\u2026..<\/p>\n<p><strong>34. Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25<sup>th<\/sup>\u00a0term.<\/strong><\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>Given,<\/p>\n<p>First term of the A.P is 1505 and<\/p>\n<p>S<sub>14<\/sub>\u00a0= 1505<\/p>\n<p>We know that, the sum of first n terms is<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2(2a + (n \u2212 1)d)<\/p>\n<p>So,<\/p>\n<p>S<sub>14<\/sub>\u00a0= 14\/2(2(10) + (14 \u2212 1)d)\u00a0= 1505<\/p>\n<p>7(20 + 13d) = 1505<\/p>\n<p>20 + 13d = 215<\/p>\n<p>13d = 215 \u2013 20<\/p>\n<p>d = 195\/13<\/p>\n<p>d =15<\/p>\n<p>Thus, the 25<sup>th<\/sup>\u00a0term is given by<\/p>\n<p>a<sub>25<\/sub>\u00a0= 10 + (25 -1)15<\/p>\n<p>= 10 + (24)15<\/p>\n<p>= 10 + 360<\/p>\n<p>= 370<\/p>\n<p>Therefore, the 25<sup>th<\/sup>\u00a0term of the A.P is 370<\/p>\n<p><strong>35. In an A.P. , the first term is 2, the last term is 29 and the sum of the terms is 155. Find the common difference of the A.P.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The first term of the A.P. (a) = 2<\/p>\n<p>The last term of the A.P. (l) = 29<\/p>\n<p>And, sum of all the terms (S<sub>n<\/sub>) = 155<\/p>\n<p>Let the common difference of the A.P. be d.<\/p>\n<p>So, find the number of terms by sum of terms formula<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2 (a + l)<\/p>\n<p>155 = n\/2(2 + 29)<\/p>\n<p>155(2) = n(31)<\/p>\n<p>31n = 310<\/p>\n<p>n = 10<\/p>\n<p>Using n for the last term, we have<\/p>\n<p>l = a + (n \u2013 1)d<\/p>\n<p>29 = 2 + (10 \u2013 1)d<\/p>\n<p>29 = 2 + (9)d<\/p>\n<p>29 \u2013 2 = 9d<\/p>\n<p>9d = 27<\/p>\n<p>d = 3<\/p>\n<p>Hence, the common difference of the A.P. is d = 3<\/p>\n<p><strong>36. The first and the last term of an A.P are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>In an A.P first term (a) = 17 and the last term (l) = 350<\/p>\n<p>And, the common difference (d) = 9<\/p>\n<p>We know that,<\/p>\n<p>a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>so,<\/p>\n<p>a<sub>n<\/sub>\u00a0= l = 17 + (n \u2013 1)9 = 350<\/p>\n<p>17 + 9n \u2013 9 = 350<\/p>\n<p>9n = 350 \u2013 8<\/p>\n<p>n = 342\/9<\/p>\n<p>n = 38<\/p>\n<p>So, the sum of all the terms of the A.P is given by<\/p>\n<p>S<sub>n<\/sub>\u00a0= n\/2 (a + l)<\/p>\n<p>= 38\/2(17 + 350)<\/p>\n<p>= 19(367)<\/p>\n<p>= 6973<\/p>\n<p>Therefore, the sum of terms of the A.P is 6973.<\/p>\n<p><strong>37. Find the number of terms of the A.P. \u201312, \u20139, \u20136, . . . , 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>First term, a\u00a0= -12<\/p>\n<p>Common difference, d = a<sub>2<\/sub>\u00a0\u2013 a<sub>1<\/sub>\u00a0= \u2013 9 \u2013\u00a0(- 12)<\/p>\n<p>d = \u2013 9 + 12 = 3<\/p>\n<p>And, we know that n<sup>th<\/sup>\u00a0term =\u00a0a<sub>n<\/sub>\u00a0= a + (n \u2013 1)d<\/p>\n<p>\u27f9 21 = -12 + (n \u2013 1)3<\/p>\n<p>\u27f9 21 = -12 + 3n \u2013 3<\/p>\n<p>\u27f9 21 = 3n \u2013 15<\/p>\n<p>\u27f9 36 = 3n<\/p>\n<p>\u27f9 n = 12<\/p>\n<p>Thus, the number of terms is 12.<\/p>\n<p>Now, if 1 is added to each of the 12 terms, the sum will increase by 12.<\/p>\n<p>Hence, the sum of all the terms of the A.P. so obtained is<\/p>\n<p>\u27f9 S<sub>12\u00a0<\/sub>+ 12 = 12\/2[a + l] + 12<\/p>\n<p>= 6[-12 + 21] + 12<\/p>\n<p>= 6 \u00d7 9 + 12<\/p>\n<p>= 66<\/p>\n<p>Therefore, the sum after adding 1 to each of the terms in the A.P is 66.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-9-exercise-96\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631608038966\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-the-rd-sharma-class-10-solutions-chapter-9-exercise-96-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 available on the Kopykitab website?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, the PDFs of RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 are available. These solutions are created in a unique method by Kopykitab\u2019s expert faculty.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631608285178\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-key-benefits-of-learning-rd-sharma-class-10-solutions-chapter-9-exercise-96\"><\/span>What are the key benefits of learning RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>The RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 answers were quite helpful. This makes it simple to clear any questions about arithmetic progression. Easily answer all of the questions that the Class 10 students have given for exercise.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631608453560\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-rd-sharma-class-10-solutions-chapter-9-exercise-96-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6:\u00a0This exercise is about problems with an A.P.&#8217;s sum of terms. RD Sharma Class 10 Solutions is the best material for students to refer to and study effectively for their exams because it covers a wide range of real problems. The following are the\u00a0RD Sharma Class 10 &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions Exercise 9.6 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-9-exercise-9-6\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions Exercise 9.6 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":127363,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/127328"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=127328"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/127328\/revisions"}],"predecessor-version":[{"id":127443,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/127328\/revisions\/127443"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/127363"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=127328"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=127328"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=127328"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}