{"id":127330,"date":"2023-09-13T14:21:00","date_gmt":"2023-09-13T08:51:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=127330"},"modified":"2023-11-16T11:07:03","modified_gmt":"2023-11-16T05:37:03","slug":"rd-sharma-class-10-solutions-chapter-9-exercise-9-4","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-9-exercise-9-4\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions Exercise 9.4 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-127361\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-9-Exercise-9.4.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 9 Exercise 9.4\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-9-Exercise-9.4.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-9-Exercise-9.4-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.4:&nbsp;<\/strong>This exercise proves the concepts of the general term, nth term from the end, and middle term of an A.P. We strongly advise students to refer to the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> provided by subject experts at Kopykitab as a very important resource for students to study well for their exams, as these are the main sections of the chapter. For a better understanding of the questions, students can use <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 9.4 PDF.<\/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-69d8b12f8c0eb\" 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\" 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Sharma Class 10 Solutions Chapter 9 Exercise 9.4 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/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-9.4.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-9.4.pdf\">RD Sharma Class 10 Solutions Chapter 9 Exercise 9.4<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-class-10-solutions-chapter-9-exercise-94-important-question-with-answers\"><\/span>Access answers to RD Sharma Class 10 Solutions Chapter 9 Exercise 9.4- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.4 Question 1.<br>(i) 10th term of the A.P. 1, 4, 7, 10, \u2026\u2026\u2026<br>(ii) 18th term of the A.P. \u221a2, 3\u221a2, 5\u221a2, \u2026\u2026\u2026.<br>(iii) nth term of the A.P. 13, 8, 3, -2, \u2026\u2026..<br>(iv) 10th term of the A.P. -40, -15, 10, 35, \u2026\u2026..<br>(v) 8th term of the A.P. 117, 104, 91, 78, \u2026\u2026\u2026..<br>(vi) 11th term of the A.P. 10.0 , 10.5, 11.0, 11.5, \u2026\u2026\u2026.<br>(vii) 9th term of the A.P. 34, 54, 74, 94, \u2026\u2026\u2026<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1740\/27766581917_068580ccbf_o.png\" alt=\"RD Sharma Class 10 Chapter 9 Arithmetic Progressions \" width=\"328\" height=\"172\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1758\/41736052675_155fb51d03_o.png\" alt=\"Arithmetic Progressions Class 10 RD Sharma \" width=\"321\" height=\"273\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1753\/27766581557_550cececce_o.png\" alt=\"RD Sharma Class 10 Chapter 9 Arithmetic Progressions \" width=\"360\" height=\"394\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1732\/27766581757_d5dcacc620_o.png\" alt=\"Arithmetic Progressions Class 10 RD Sharma \" width=\"363\" height=\"650\"><\/p>\n<p>Question 2.<br>(i) Which term of the A.P. 3, 8, 13, \u2026\u2026 is 248?<br>(ii) Which term of the A.P. 84, 80, 76, \u2026.. is 0?<br>(iii) Which term of the A.P. 4, 9, 14, \u2026.. is 254?<br>(iv) Which term of the A.P. 21, 42, 63, 84, \u2026.. is 420?<\/p>\n<p><br>(v) Which term of the A.P. 121, 117, 113, \u2026.. is its first negative term?<br>Solution:<br>(i) A.P. is 3, 8, 13, \u2026, 248<br>Here first term (a) = 3<br>and common difference (d) = 8 \u2013 3 = 5<br><img src=\"https:\/\/farm2.staticflickr.com\/1723\/27766582847_2a836028f1_o.png\" alt=\"RD Sharma Class 10 Solutions Arithmetic Progressions \" width=\"352\" height=\"578\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1745\/27766582267_ec88b95423_o.png\" alt=\"RD Sharma Class 10 Solutions Arithmetic Progressions Exercise 9.4 \" width=\"358\" height=\"395\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1759\/27766582507_94cd7d8f87_o.png\" alt=\"RD Sharma Class 10 Solutions Arithmetic Progressions \" width=\"360\" height=\"432\"><\/p>\n<p>Question 3.<br>(i) Is 68 a term of the A.P. 7, 10, 13, \u2026\u2026?<br>(ii) Is 302 a term of the A.P. 3, 8, 13, \u2026..?<br>(ii) Is -150 a term of the A.P. 11, 8, 5, 2, \u2026\u2026?<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1756\/27766583607_eab344ba2d_o.png\" alt=\"RD Sharma Class 10 Solutions Arithmetic Progressions Exercise 9.4 \" width=\"349\" height=\"451\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1749\/27766583367_d25aea60b0_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions \" width=\"358\" height=\"331\"><\/p>\n<p>Question 4.<br>How many terms are there in the A.P.?<br>(i) 7, 10, 13, \u2026 43<br>(ii) -1, \u2013&nbsp;56&nbsp;, \u2013&nbsp;23&nbsp;, \u2013&nbsp;12&nbsp;, \u2026\u2026..,&nbsp;103<br>(iii) 7, 13, 19, \u2026, 205<br>(iv) 18, 1512&nbsp;, 13, \u2026, -47<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1747\/27766584577_bdffc76fd9_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 9 Arithmetic Progressions \" width=\"351\" height=\"429\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1736\/27766583757_59ee486a70_o.png\" alt=\"RD Sharma Solutions Class 10 Chapter 9 Arithmetic Progressions \" width=\"359\" height=\"553\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1724\/27766583967_f3fede5d0f_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 9 Arithmetic Progressions \" width=\"364\" height=\"420\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1731\/27766584257_ac585d7ea4_o.png\" alt=\"RD Sharma Solutions Class 10 Chapter 9 Arithmetic Progressions \" width=\"227\" height=\"420\"><\/p>\n<p>Question 5.<br>The first term of an A.P. is 5, the common difference is 3 and the last term is 80; find the number of terms.<br>Solution:<br>The first term of an A.P. (a) = 5<br>and common difference (d) = 3<br>Last term = 80<br>Let the last term be nth<br>a<sub>n<\/sub>&nbsp;= a + (n \u2013 1) d<br>\u21d2 80 = 5 + (n \u2013 1) x 3<br>\u21d2 80= 5 + 3n \u2013 3<br>\u21d2 3n = 80 \u2013 5 + 3 = 78<br>\u21d2 n = 26<br>Number of terms = 26<\/p>\n<p>Question 6.<br>The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.<br>Solution:<br>6th term of A.P. = 19<br>and 17th term = 41<br>Let a be the first term, and d be the common difference<br>We know that<br><img src=\"https:\/\/farm2.staticflickr.com\/1734\/27766584657_a4280e7f54_o.png\" alt=\"Learncbse.In Class 10 Chapter 9 Arithmetic Progressions \" width=\"350\" height=\"393\"><\/p>\n<p>Question 7.<br>If the 9th term of an A.P. is zero, prove that its 29th term is double the 19th term.<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1724\/27766584867_8b93ea6820_o.png\" alt=\"Class 10 RD Sharma Solutions Chapter 9 Arithmetic Progressions \" width=\"356\" height=\"331\"><\/p>\n<p>Question 8.<br>If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that the 25th term of the A.P. is zero.<br>Solution:<br>Let a, a + d, a + 2d, a + 3d, \u2026\u2026\u2026 be an A.P.<br>a<sub>n<\/sub>&nbsp;= a + (n \u2013 1) d<br>Now a<sub>10<\/sub>&nbsp;= a + (10 \u2013 1) d = a + 9d<br>and a<sub>15<\/sub>&nbsp;= a + (15 \u2013 1) d = a + 14d<br><img src=\"https:\/\/farm2.staticflickr.com\/1726\/27766585067_524c39ab77_o.png\" alt=\"Class 10 RD Sharma Solutions Chapter 9 Arithmetic Progressions \" width=\"353\" height=\"283\"><\/p>\n<p>Question 9.<br>The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1731\/27766585217_1c4e97d4a5_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 9 Arithmetic Progressions \" width=\"353\" height=\"375\"><\/p>\n<p>Question 10.<br>In a certain A.P., the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.<br>Solution:<br>Let a, a + d, a + 2d, a + 3d, \u2026\u2026.. be an A.P.<br>a<sub>n<\/sub>&nbsp;= a + (n \u2013 1) d<br>10th (a<sub>10<\/sub>) = a + (10 \u2013 1) d = a + 9d<br>and 24th term (a<sub>24<\/sub>) = a + (24 \u2013 1) d = a + 23d<br>24th term = 2 x 10th term<br>a + 23d = 2 (a + 9d)<br>\u21d2 a + 23d = 2a + 18d<br>\u21d2 2a \u2013 a = 23d \u2013 18d<br>\u21d2 a = 5d \u2026.(i)<br>Now 72nd term = a + (72 \u2013 1)d = a + 71d<br>and 34th term = a + (34 \u2013 1) d = a + 33d<br>Now a + 71d \u2013 5d + 71d = 76d<br>and a + 33d = 5d+ 33d = 38d<br>76d = 2 x 38d<br>72th term = 2 (34th term) = twice of the 34th term<br>Hence proved.<\/p>\n<p>Question 11.<br>The 26th, 11th, and last terms of an A.P. are 0, 3, and \u2013 15, respectively. Find the common difference and the number of terms. [NCERT Exemplar]<br>Solution:<br>Let the first term, common difference, and a number of terms of an A.P. are a, d and n, respectively.<br>We know that, if the last term of an A.P. is known, then<br>l = a + (n \u2013 1) d \u2026\u2026(i)<br>and nth term of an A.P is<br><img src=\"https:\/\/farm2.staticflickr.com\/1758\/27766585397_dcc029f724_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 9 Arithmetic Progressions \" width=\"355\" height=\"385\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1741\/27766585297_edfd8eeeb9_o.png\" alt=\"RD Sharma Class 10 Solution Chapter 9 Arithmetic Progressions \" width=\"350\" height=\"584\"><\/p>\n<p>Question 12.<br>If the nth term of the A.P. 9, 7, 5, \u2026 is the same as the nth term of the A.P. 15, 12, 9, \u2026 find n.<br>Solution:<br>In A.P 9, 7, 5, \u2026\u2026\u2026<br>Here first term (a) = 9 and d = 7 \u2013 9 = -2 {or 5 \u2013 7 = -2}<br>nth term (a<sub>n<\/sub>) = a + (n \u2013 1) d = 9 + (n \u2013 1) (-2) = 9 \u2013 2n + 2 = 11 \u2013 2n<br>Now in A.P. 15, 12, 9, \u2026..<br>Here first term (a) = 15 and (d) = 12 \u2013 15 = -3<br>nth term (a<sub>n<\/sub>) = a + (n \u2013 1) d = 15 + (n \u2013 1) x (-3)<br>The nth term of first A.P. = nth term of second A.P.<br>11 \u2013 2n = 18 \u2013 3n<br>\u21d2 -2n + 3n = 18 \u2013 11<br>\u21d2 n = 7<br>Hence n = 7<\/p>\n<p>Question 13.<br>Find the 12th term from the end of the following arithmetic progressions :<br>(i) 3, 5, 7, 9, \u2026 201<br>(ii) 3, 8, 13,\u2026, 253<br>(iii) 1, 4, 7, 10, \u2026, 88<br>Solution:<br>(i) In the A.P. 3, 5, 7, 9, \u2026 201<br>First term (a) = 3, last term (l) = 201<br>and common difference (d) = 5 \u2013 3 = 2<br>We know that nth term from the last = l \u2013 (n \u2013 1 ) d<br>12th term from the last = 201 \u2013 (12 \u2013 1) x 2 = 201 \u2013 11 x 2 = 201 \u2013 22 = 179<br>(ii) In the A.P. 3, 8, 13, \u2026, 253<br>First term (a) = 3<br>Common difference (d) = 8 \u2013 3 = 5<br>and last term = 253<br>The nth term from the last = l \u2013 (n \u2013 1) d<br>12th term from the last = 253 \u2013 (12 \u2013 1) x 5 = 253 \u2013 11 x 5 = 253 \u2013 55 = 198<br>(iii) In the A.P. 1, 4, 7, 10, \u2026, 88<br>First term (a) = 1<br>Common difference (d) = 4 \u2013 1 = 3<br>and last term = 88<br>The nth term from the last = l \u2013 (n \u2013 1) d<br>12th term from the last = 88 \u2013 (12 \u2013 1) x 3 = 88 \u2013 11 x 3 = 88 \u2013 33 = 55<\/p>\n<p>Question 14.<br>The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1735\/27766585587_c5c63823d9_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 9 Arithmetic Progressions \" width=\"352\" height=\"602\"><\/p>\n<p>Question 15.<br>Find the second term and nth term of an A.P. whose 6th term is 12 and the 8th term is 22.<br>Solution:<br>In an A.P.<br>6th term (a<sub>6<\/sub>) = 12<br>and 8th term (a<sub>8<\/sub>) = 22<br>Let a be the first term and d be a common difference, then<br><img src=\"https:\/\/farm2.staticflickr.com\/1721\/27766585707_400f131742_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 9 Arithmetic Progressions \" width=\"345\" height=\"415\"><\/p>\n<p>Question 16.<br>How many numbers of two-digit are divisible by 3?<br>Solution:<br>Let n be the number of terms which are divisible by 3 and d are of two digit numbers<br>Let a be the first term and d be the common difference, then<br>a = 12, d = 3, last term = 99<br>a<sub>n<\/sub>&nbsp;= a + (n \u2013 1) d<br>99 = 12 + (n \u2013 1) x 3<br>\u21d2 99 = 12 + 3n \u2013 3<br>\u21d2 3n = 99 \u2013 9<br>\u21d2 n = 30<br>Number of terms = 30<\/p>\n<p>Question 17.<br>An A.P. consists of 60 terms. If the first and the last terms are 7 and 125 respectively, find the 32nd term.<br>Solution:<br>In an A.P.<br>n = 60<br>First term (a) = 7 and last term (l) = 125<br>Let d be the common difference, then<br>a<sub>60<\/sub>&nbsp;= a + (60 \u2013 1) d<br>\u21d2 125 = 7 + 59d<br>\u21d2 59d = 125 \u2013 7 = 118<br>Common difference = 2<br>Now 32nd term (a<sub>32<\/sub>) = a + (32 \u2013 1) d = 7 + 31 x 2 = 7+ 62 = 69<\/p>\n<p>Question 18.<br>The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P.<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1735\/27766585857_e3fb7944b3_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 9 Arithmetic Progressions \" width=\"351\" height=\"598\"><\/p>\n<p>Question 19.<br>The first term of an A.P. is 5 and its 100th term is -292. Find the 50th term of this A.P.<br>Solution:<br>First term of an A.P. = 5<br>and 100th term = -292<br><img src=\"https:\/\/farm2.staticflickr.com\/1735\/27766585967_37b0d5d9cf_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 9 Arithmetic Progressions \" width=\"353\" height=\"268\"><\/p>\n<p>Question 20.<br>Find a<sub>30<\/sub>&nbsp;\u2013 a<sub>20<\/sub>&nbsp;for the A.P.<br>(i) -9, -14, -19, -24, \u2026<br>(ii) a, a + d, a + 2d, a + 3d, \u2026<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1744\/27766586127_c6d20da819_o.png\" alt=\"Class 10 RD Sharma Chapter 9 Arithmetic Progressions \" width=\"352\" height=\"404\"><\/p>\n<p>Question 21.<br>Write the expression a<sub>n<\/sub>&nbsp;\u2013 a<sub>k<\/sub>&nbsp;for the A.P. a, a + d, a + 2d, \u2026\u2026<br>Hence, find the common difference of the A.P. for which<br>(i) 11th term is 5 and 13th term is 79.<br>(ii) a<sub>10<\/sub>&nbsp;\u2013 a<sub>5<\/sub>&nbsp;= 200<br>(iii) 20th term is 10 more than the 18th term.<br>Solution:<br>In the A.P. a, a + d, a + 2d, \u2026..<br><img src=\"https:\/\/farm2.staticflickr.com\/1744\/27766586287_fddbbda742_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 9 Arithmetic Progressions \" width=\"353\" height=\"594\"><\/p>\n<p>Question 22.<br>Find n if the given value of x is the nth term of the given A.P.<br><img src=\"https:\/\/farm2.staticflickr.com\/1732\/27766586987_923664fdd6_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 9 Arithmetic Progressions \" width=\"276\" height=\"169\"><br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1723\/27766586387_f1d016872d_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 9 Arithmetic Progressions \" width=\"334\" height=\"151\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1753\/27766586677_70a77667c7_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 9 Arithmetic Progressions \" width=\"340\" height=\"578\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1748\/27766586787_056cb2d27c_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 9 Arithmetic Progressions \" width=\"340\" height=\"399\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1732\/27766586867_8394b71921_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions \" width=\"354\" height=\"232\"><\/p>\n<p>Question 23.<br>The eighth term of an A.P. is half of its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the 15th term.<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1730\/27766587207_6fef5b8112_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions \" width=\"352\" height=\"490\"><\/p>\n<p>Question 24.<br>Find the arithmetic progression whose third term is 16 and the seventh term exceeds its fifth term by 12.<br>Solution:<br>Let a, a + d, a + 2d, a + 3d, \u2026\u2026\u2026. be the A.P.<br>a<sub>n<\/sub>&nbsp;= a + (n \u2013 1) d<br>But a<sub>3<\/sub>&nbsp;= 16<br><img src=\"https:\/\/farm2.staticflickr.com\/1746\/27766587277_6986400f6e_o.png\" alt=\"10th Maths Solution Book Pdf Chapter 9 Arithmetic Progressions \" width=\"281\" height=\"259\"><\/p>\n<p>Question 25.<br>The 7th term of an A.P. is 32 and its 13th term is 62. Find the A.P. [CBSE 2004]<br>Solution:<br>Let a, a + d, a + 2d, a + 3d, be the A.P.<br>Here a is the first term and d is the common difference<br>a<sub>n<\/sub>&nbsp;= a + (n \u2013 1) d<br>Now a<sub>7<\/sub>&nbsp;= a + (7 \u2013 1) d = a + 6d = 32 \u2026.(i)<br>and a<sub>13<\/sub>&nbsp;= a + (13 \u2013 1) d = a + 12d = 62 \u2026.(ii)<br>Subtracting (i) from (ii)<br>6d = 30<br>\u21d2 d = 5<br>a + 6 x 5 = 32<br>\u21d2 a + 30 = 32<br>\u21d2 a = 32 \u2013 30 = 2<br>A.P. will be 2, 7, 12, 17, \u2026\u2026\u2026..<\/p>\n<p>Question 26.<br>Which term of the A.P. 3, 10, 17, \u2026 will be 84 more than its 13th term? [CBSE 2004]<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1754\/27766587547_b4d6cce43b_o.png\" alt=\"RD Sharma 10 Chapter 9 Arithmetic Progressions \" width=\"351\" height=\"279\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1726\/27766587357_872a713f96_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 9 Arithmetic Progressions \" width=\"194\" height=\"116\"><\/p>\n<p>Question 27.<br>Two arithmetic progressions have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1728\/42635710881_305ef5c2cd_o.png\" alt=\"RD Sharma 10 Chapter 9 Arithmetic Progressions \" width=\"349\" height=\"354\"><\/p>\n<p>Question 28.<br>For what value of n, the nth terms of the arithmetic progressions 63, 65, 67,\u2026 and 3, 10, 17, \u2026 are equal? (C.B.S.E. 2008)<br>Solution:<br>In the A.P. 63, 65, 67, \u2026<br>a = 63 and d = 65 \u2013 63 = 2<br>a<sub>n<\/sub>&nbsp;= a<sub>1<\/sub>&nbsp;+ (n \u2013 1) d = 63 + (n \u2013 1) x 2 = 63 + 2n \u2013 2 = 61 + 2n<br>and in the A.P. 3, 10, 17, \u2026<br>a = 3 and d = 10 \u2013 3 = 7<br>a<sub>n<\/sub>&nbsp;= a + (n \u2013 1) d = 3 + (n \u2013 1) x 7 = 3 + 7n \u2013 7 = 7n \u2013 4<br>But both nth terms are equal<br>61 + 2n = 7n \u2013 4<br>\u21d2 61 + 4 = 7n \u2013 2n<br>\u21d2 65 = 5n<br>\u21d2 n = 13<br>n = 13<\/p>\n<p>Question 29.<br>How many multiples of 4 lie between 10 and 250?<br>Solution:<br>All the terms between 10 and 250 are multiple of 4<br>First multiple (a) = 12<br>and last multiple (l) = 248<br>and d = 4<br>Let n be the number of multiples, then<br>a<sub>n<\/sub>&nbsp;= a + (n \u2013 1) d<br>\u21d2 248 = 12 + (n \u2013 1) x 4 = 12 + 4n \u2013 4<br>\u21d2 248 = 8 + 4n<br>\u21d2 4n = 248 \u2013 8 = 240<br>n = 60<br>Number of terms are = 60<\/p>\n<p>Question 30.<br>How many three-digit numbers are divisible by 7?<br>Solution:<br>First three digit number is 100 and last three digit number is 999<br>In the sequence of the required three digit numbers which are divisible by 7, will be between<br>a = 105 and last number l = 994 and d = 7<br>Let n be the number of terms, then<br>a<sub>n<\/sub>&nbsp;= a + (n \u2013 1) d<br>994 = 105 + (n \u2013 1) x 7<br>994 = 105 + 7n \u2013 7<br>\u21d2 7n = 994 \u2013 105 + 7<br>\u21d2 7n = 896<br>\u21d2 n = 128<br>Number of terms =128<\/p>\n<p>Question 31.<br>Which term of the arithmetic progression 8, 14, 20, 26, \u2026 will be 72 more than its 41st term? (C.B.S.E. 2006C)<br>Solution:<br>In the given A.P. 8, 14, 20, 26, \u2026<br><img src=\"https:\/\/farm2.staticflickr.com\/1747\/42635710951_41c8e4c7cc_o.png\" alt=\"Solution Of RD Sharma Class 10 Chapter 9 Arithmetic Progressions \" width=\"296\" height=\"323\"><\/p>\n<p>Question 32.<br>Find the term of the arithmetic progression 9, 12, 15, 18, \u2026 which is 39 more than its 36th term (C.B.S.E. 2006C)<br>Solution:<br>In the given A.R 9, 12, 15, 18, \u2026<br>First term (a) = 9<br>and common difference (d) = 12 \u2013 9 = 3<br>and a<sub>n<\/sub>&nbsp;= a + (n \u2013 1) d<br>Now a<sub>36<\/sub>&nbsp;= a + (36 \u2013 1) d = 9 + 35 x 3 = 9 + 105 = 114<br>Let the an be the required term<br>a<sub>n<\/sub>&nbsp;= a + (n \u2013 1) d<br>= 9 + (n \u2013 1) x 3 = 9 + 3n \u2013 3 = 6 + 3n<br>But their difference is 39<br>a<sub>n<\/sub>&nbsp;\u2013 a<sub>36<\/sub>&nbsp;= 39<br>\u21d2 6 + 3n \u2013 114 = 39<br>\u21d2 114 \u2013 6 + 39 = 3n<br>\u21d2 3n = 147<br>\u21d2 n = 49<br>Required term is 49th<\/p>\n<p>Question 33.<br>Find the 8th term from the end of the A.P. 7, 10, 13, \u2026, 184. (C.B.S.E. 2005)<br>Solution:<br>The given A.P. is 7, 10, 13,\u2026, 184<br>Here first term (a) = 7<br>and common difference (d) = 10 \u2013 7 = 3<br>and last tenn (l) = 184<br>Let nth term from the last is a<sub>n<\/sub>&nbsp;= l \u2013 (n \u2013 1) d<br>a<sub>8<\/sub>= 184 \u2013 (8 \u2013 1) x 3 = 184 \u2013 7 x 3 = 184 \u2013 21 = 163<\/p>\n<p>Question 34.<br>Find the 10th term from the end of the A.P. 8, 10, 12, \u2026, 126. (C.B.S.E. 2006)<br>Solution:<br>The given A.P. is 8, 10, 12, \u2026, 126<br>Here first term (a) = 8<br>Common difference (d) = 10 \u2013 8 = 2<br>and last tenn (l) = 126<br>Now nth term from the last is a<sub>n<\/sub>&nbsp;= l \u2013 (n \u2013 1) d<br>a<sub>10<\/sub>&nbsp;= 126 \u2013 (10 \u2013 1) x 2 = 126 \u2013 9 x 2 = 126 \u2013 18 = 108<\/p>\n<p>Question 35.<br>The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the A.P. (C.B.S.E. 2009)<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1722\/42635711091_1c595e49a7_o.png\" alt=\"Solution Of RD Sharma Class 10 Chapter 9 Arithmetic Progressions \" width=\"357\" height=\"461\"><\/p>\n<p>Question 36.<br>Which term of the A.P. 3, 15, 27, 39, \u2026. will be 120 more than its 21st term? (C.B.S.E. 2009)<br>Solution:<br>A.P. is given : 3, 15, 27, 39, \u2026\u2026.<br>Here first term (a) = 3<br>and c.d. (d) = 15 \u2013 3 = 12<br>Let nth term be the required term<br>Now 21st term = a + (n \u2013 1) d = 3 + 20 x 12 = 3 + 240 = 243<br>According to the given condition,<br>nth term \u2013 21 st term = 120<br>\u21d2 a + (n \u2013 1) d \u2013 243 = 120<br>\u21d2 3 + (n \u2013 1) x 12 = 120 + 243 = 363<br>\u21d2 (n \u2013 1) 12 = 363 \u2013 3 = 360<br>\u21d2 n \u2013 1 = 30<br>\u21d2 n = 30 + 1 = 31<br>31 st term is the required term<\/p>\n<p>Question 37.<br>The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, find the nth term.[CBSE 2012]<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1756\/42635711191_a91500b8c9_o.png\" alt=\"RD Sharma 10 Solutions Chapter 9 Arithmetic Progressions \" width=\"364\" height=\"390\"><\/p>\n<p>Question 38.<br>Find the number of ail three digit natural numbers which are divisible by 9. [CBSE 2013]<br>Solution:<br>First 3-digit number which is divisible by 9 = 108<br>and last 3-digit number = 999<br>d= 9<br>a + (n \u2013 1) d = 999<br>\u21d2 108 + (n \u2013 1) x 9 = 999<br>\u21d2 (n \u2013 1) d = 999 \u2013 108<br>\u21d2 (n \u2013 1) x 9 = 891<br>\u21d2 n \u2013 1 = 99<br>\u21d2 n = 99 + 1 = 100<br>Number of terms = 100<\/p>\n<p>Question 39.<br>The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P. [CBSE 2013]<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1726\/42635711321_1c37e9edf4_o.png\" alt=\"RD Sharma Maths Book For Class 10 Solution Chapter 9 Arithmetic Progressions \" width=\"361\" height=\"555\"><\/p>\n<p>Question 40.<br>The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P. [CBSE 2013]<br>Solution:<br>Let a be the first term and d be the common difference and<br>T<sub>n<\/sub>&nbsp;= a + (n \u2013 1) d<br><img src=\"https:\/\/farm2.staticflickr.com\/1746\/42635711451_730252326b_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 9 Arithmetic Progressions \" width=\"354\" height=\"519\"><\/p>\n<p>Question 41.<br>The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is 4 times its 15th term. [CBSE 2013]<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1737\/42635711581_03b945bd02_o.png\" alt=\"RD Sharma Mathematics Class 10 Pdf Download Free Chapter 9 Arithmetic Progressions \" width=\"355\" height=\"349\"><br>Hence 72nd term = 4 times of 15th term<\/p>\n<p>Question 42.<br>Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5. [CBSE 2014]<br>Solution:<br>Numbers divisible by both 2 and 5 are 110, 120, 130, \u2026\u2026\u2026. , 990<br>Here a = 110, x = 120 \u2013 110 = 10<br>a<sub>n<\/sub>&nbsp;= 990<br>As a + (n \u2013 1) d = 990<br>110 + (n \u2013 1) (10) = 990<br>(n \u2013 1) (10) = 990 \u2013 110 = 880<br>n \u2013 1 = 88<br>n = 88 + 1 = 89<\/p>\n<p>Question 43.<br>If the seventh term of an AP is 19 and its ninth term is 17, find its (63) rd term. [CBSE 2014]<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1721\/42635711791_8fcbc4ba13_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 9 Arithmetic Progressions \" width=\"344\" height=\"428\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1726\/42635711701_83bee80cca_o.png\" alt=\"Class 10 RD Sharma Pdf Chapter 9 Arithmetic Progressions \" width=\"278\" height=\"161\"><\/p>\n<p>Question 44.<br>The sum of the 5th and 9th terms of an AP is 30. If its 25th term is three times its 8th term, find the AP. [CBSE 2014]<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1726\/42635712001_02461873e0_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 9 Arithmetic Progressions \" width=\"340\" height=\"521\"><\/p>\n<p>Question 45.<br>Find where 0 (zero) is a term of the AP 40, 37, 34, 31, \u2026\u2026 [CBSE 2014]<br>Solution:<br>AP 40, 37, 34, 31, \u2026..<br>Here a = 40, d = -3<br>Let T<sub>n<\/sub>&nbsp;= 0<br>T<sub>n<\/sub>&nbsp;= a + (n \u2013 1) d<br>\u21d2 0 = 40 + (n \u2013 1) (-3)<br>\u21d2 0 = 40 \u2013 3n + 3<br>\u21d2 3n = 43<br>\u21d2 n =&nbsp;433&nbsp;which is in fraction<br>There is no term which is 0<\/p>\n<p>Question 46.<br>Find the middle term of the A.P. 213, 205, 197, \u2026, 37. [CBSE2015]<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1757\/42635712061_8f933a1967_o.png\" alt=\"Class 10 RD Sharma Pdf Chapter 9 Arithmetic Progressions \" width=\"356\" height=\"425\"><\/p>\n<p>Question 47.<br>If the 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term, find the A.P. [BTE2015]<br>Solution:<br>We know that,<br>T<sub>n<\/sub>&nbsp;= a + (n \u2013 1 )d<br>T<sub>5<\/sub>&nbsp;= a + 4d \u21d2 a + 4d = 31 \u2026\u2026(i)<br>and T<sub>25<\/sub>&nbsp;= a + 24d<br>\u21d2a + 24d = 140 + T<sub>5<\/sub><br>\u21d2 a + 24d = 140 + 31 = 171 \u2026..(ii)<br>Subtracting (i) from (ii),<br>20d= 140<br>and a + 4d = 31<br>\u21d2 a + 4 x 7 = 31<br>\u21d2 a + 28 = 31<br>\u21d2 a = 31 \u2013 28 = 3<br>a = 3 and d = 7<br>AP will be 3, 10, 17, 24, 31, \u2026\u2026..<\/p>\n<p>Question 48.<br>Find the sum of two middle terms of the<br><img src=\"https:\/\/farm2.staticflickr.com\/1734\/42635712481_d26192aa09_o.png\" alt=\"RD Sharma Class 10 Solutions Pdf Free Download Chapter 9 Arithmetic Progressions \" width=\"329\" height=\"80\"><br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1725\/42635712181_ca7f5a1490_o.png\" alt=\"RD Sharma Class 10 Solutions Arithmetic Progressions \" width=\"356\" height=\"328\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1725\/42635712311_dfaa1a5043_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions \" width=\"358\" height=\"353\"><\/p>\n<p>Question 49.<br>If (m + 1)th term of an A.P. is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n + 1)th term.<br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1752\/42635712561_4251fdc1e7_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 9 Arithmetic Progressions \" width=\"355\" height=\"518\"><\/p>\n<p>Question 50.<br>If an A.P. consists of n terms with first term a and nth term l show that the sum of the mth term from the beginning and the mth term from the end is (a + l).<br>Solution:<br>In an A.P.<br>Number of terms = n<br>First term = a<br>and nth term = l<br>mth term (a<sub>m<\/sub>) = a + (m \u2013 1) d<br>and mth term from the end = l \u2013 (m \u2013 1)d<br>Their sum = a + (m \u2013 1) d + l \u2013 (m \u2013 1) d = a + l<br>Hence proved.<\/p>\n<p>Question 51.<br>How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3? [NCERT Exemplar]<br>Solution:<br>Here, the first number is 11, which divided by 4 leave remainder 3 between 10 and 300.<br>Last term before 300 is 299, which divided by 4 leave remainder 3.<br>11, 15, 19, 23, \u2026, 299<br>Here, first term (a) = 11,<br>common difference (d) = 15 \u2013 11 = 4<br>nth term, a<sub>n<\/sub>&nbsp;= a + (n \u2013 1 ) d = l [last term]<br>\u21d2 299 = 11 + (n \u2013 1) 4<br>\u21d2 299 \u2013 11 = (n \u2013 1) 4<br>\u21d2 4(n \u2013 1) = 288<br>\u21d2 (n \u2013 1) = 72<br>n = 73<\/p>\n<p>Question 52.<br>Find the 12th term from the end of the A.P. -2, -4, -6, \u2026, -100. [NCERT Exemplar]<br>Solution:<br>Given, A.P., -2, -4, -6, \u2026, -100<br>Here, first term (a) = -2,<br>common difference (d) = -4 \u2013 (-2)<br>and the last term (l) = -100.<br>We know that, the nth term an of an A.P. from the end is a<sub>n<\/sub>&nbsp;= l \u2013 (n \u2013 1 )d,<br>where l is the last term and d is the common difference. 12th term from the end,<br>a<sub>n<\/sub>&nbsp;= -100 \u2013 (12 \u2013 1) (-2)<br>= -100 + (11) (2) = -100 + 22 = -78<br>Hence, the 12th term from the end is -78<\/p>\n<p>Question 53.<br>For the A.P.: -3, -7, -11,\u2026, can we find a<sub>30<\/sub>&nbsp;\u2013 a<sub>20<\/sub>&nbsp;without actually finding a<sub>30<\/sub>&nbsp;and a<sub>20<\/sub>? Give reasons for your answer. [NCERT Exemplar]<br>Solution:<br>True.<br>nth term of an A.P., a<sub>n<\/sub>&nbsp;= a + (n \u2013 1)d<br>a<sub>30<\/sub>&nbsp;= a + (30 \u2013 1 )d = a + 29d<br>and a<sub>20<\/sub>&nbsp;= a + (20 \u2013 1 )d = a + 19d \u2026(i)<br>Now, a<sub>30<\/sub>&nbsp;\u2013 a<sub>20<\/sub>&nbsp;= (a + 29d) \u2013 (a + 19d) = 10d<br>and from given A.P.<br>common difference, d = -7 \u2013 (-3) = -7 + 3 = -4<br>a<sub>30<\/sub>&nbsp;\u2013 a<sub>20&nbsp;<\/sub>= 10(-4) = -40 [from Eq- (i)]<\/p>\n<p>Question 54.<br>Two A.P.s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why? [NCERT Exemplar]<br>Solution:<br>Let the same common difference of two A.P.\u2019s is d.<br>Given that, the first term of first A.P. and second A.P. are 2 and 7 respectively,<br>then the A.P.\u2019s are 2, 2 + d, 2 + 2d, 2 + 3d, \u2026 and 7, 7 + d, 7 + 2d, 7 + 3d, \u2026<br>Now, 10th terms of first and second A.P.\u2019s are 2 + 9d and 7 + 9d, respectively.<br>So, their difference is 7 + 9d \u2013 (2 + 9d) = 5<br>Also, 21st terms of first and second A.P.\u2019s are 2 + 20d and 7 + 20d, respectively.<br>So, their difference is 7 + 20d \u2013 (2 + 9d) = 5<br>Also, if the a<sub>n<\/sub>&nbsp;and b<sub>n<\/sub>&nbsp;are the nth terms of first and second A.P.<br>Then b<sub>n<\/sub>&nbsp;\u2013 a<sub>n<\/sub>&nbsp;= [7 + (n \u2013 1 ) d] \u2013 [2 + (n \u2013 1) d = 5<br>Hence, the difference between any two corresponding terms of such A.P.\u2019s is the same as the difference between their first terms.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 9 Exercise 9.4. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>&nbsp;Class 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-94\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 9 Exercise 9.4<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-1631607738859\" 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-94-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 9 Exercise 9.4 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 9.4 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-1631608272188\" 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-94\"><\/span>What are the key benefits of learning RD Sharma Class 10 Solutions Chapter 9 Exercise 9.4?<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 9.4 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-1631608444934\" 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-94-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 9 Exercise 9.4 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 9.4 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 9.4:&nbsp;This exercise proves the concepts of the general term, nth term from the end, and middle term of an A.P. We strongly advise students to refer to the RD Sharma Class 10 Solutions provided by subject experts at Kopykitab as a very important resource for students to &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions Exercise 9.4 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-9-exercise-9-4\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions Exercise 9.4 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":127361,"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\/127330"}],"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=127330"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/127330\/revisions"}],"predecessor-version":[{"id":507985,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/127330\/revisions\/507985"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/127361"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=127330"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=127330"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=127330"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}