{"id":47564,"date":"2023-08-31T09:00:00","date_gmt":"2023-08-31T03:30:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=47564"},"modified":"2023-12-13T11:17:57","modified_gmt":"2023-12-13T05:47:57","slug":"rs-aggarwal-solutions-class-10-maths-chapter-11-arithmetic-progression","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-solutions-class-10-maths-chapter-11-arithmetic-progression\/","title":{"rendered":"RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression | Updated For 2024"},"content":{"rendered":"\n<p><img class=\"alignnone wp-image-135331 size-full\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Solutions-Class-10-Maths-Chapter-11.jpg\" alt=\"RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Solutions-Class-10-Maths-Chapter-11.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Solutions-Class-10-Maths-Chapter-11-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression: <\/strong>Download the Free PDF of RS Aggarwal Solutions Class 10 Maths Chapter 11 from here. All the solutions of <a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-10-maths-rs-aggarwal-solutions\/\" target=\"_blank\" rel=\"noopener\">RS Aggarwal Solutions Class 10 Maths<\/a> are prepared by subject matter experts and are as per the current CBSE Syllabus. RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression is a great help guide for you.<\/p>\n<p>You can access the RS Aggarwal Solutions Class 10 Maths Chapter 11 PDF offline as well. For more details about the RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression, read the whole blog.<\/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-69e182af2d5db\" 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 d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" 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Chapter 11 Arithmetic Progression PDF?\">From where can I find the download link for the RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-solutions-class-10-maths-chapter-11-arithmetic-progression\/#how-much-does-it-cost-to-download-the-rs-aggarwal-solutions-class-10-maths-chapter-11-arithmetic-progression-pdf\" title=\"How much does it cost to download the\u00a0RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression PDF?\">How much does it cost to download the\u00a0RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" 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Progression?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-the-free-pdf-of-rs-aggarwal-solutions-class-10-maths-chapter-11-arithmetic-progression\"><\/span>Download the Free PDF Of RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Solutions-for-Class-10-Maths-Chapter-11-Arithmetic-Progression.pdf\" target=\"_blank\" rel=\"noopener\">RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression<\/a><\/p>\n<p>&nbsp;<\/p>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<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\/10\/RS-Aggarwal-Solutions-for-Class-10-Maths-Chapter-11-Arithmetic-Progression.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-solutions-for-rs-aggarwal-solutions-class-10-maths-chapter-11\"><\/span>Access Solutions For RS Aggarwal Solutions Class 10 Maths Chapter 11<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1:<\/strong><br>The given progression is 3, 9, 15, 21 \u2026..<br>Clearly (9 \u2013 3) = (15 \u2013 9) = (21 \u2013 15) = 6 which is constant<br>Thus, each term differs from its preceding term by 6<br>So, the given progression is an AP<br>Its first term = 3 and the common difference = 6<\/p>\n<p><strong>Question 2:<\/strong><br>The given progression is 16, 11, 6, 1, -4 \u2026.<br>Clearly (11 \u2013 16) = (1 \u2013 6) = (-4 \u2013 1) = \u2013 5 which is constant<br>Thus, each term differs from its preceding term by \u2013 5<br>So the given progression is an AP<br>Its first term = 16 and the common difference = \u2013 5<br><strong>Question 3:<\/strong><br>(i) The given AP is 1, 5, 9, 13, 17\u2026..<br>Its first term = 1 and common difference = (5 \u2013 1) = 4<br>\u2234&nbsp;a = 1 and d = 4<br>The n<sup>th<\/sup>&nbsp;term of the AP is given by<br>T<sub>n&nbsp;<\/sub>= a + (n-1) d<br>T<sub>20<\/sub>&nbsp;= 1 + (20-1) x 4 = 1+ 76 = 77<br>Hence, the 20<sup>th<\/sup>&nbsp;term is 77<br>(ii) The given AP is 6, 9, 12, 15 \u2026\u2026<br>Its first term = 6 and common difference = (9 \u2013 6) = 3<br>\u2234&nbsp;a = 6, d = 3<br>The n<sup>th<\/sup>&nbsp;term of the AP is given by<br>T<sub>n&nbsp;<\/sub>= a + (n-1) d<br>T35&nbsp;= 6 + (35-1) x 3 = 6+ 102 = 108<br>Hence, the 35<sup>th<\/sup>&nbsp;term is 108<br>(iii) The given AP is 5, 11, 17, 23 \u2026..<br>Its first term = 5, and common difference = (11 \u2013 5) = 6<br>\u2234&nbsp;a = 5, d = 6<br>The n<sup>th<\/sup>&nbsp;term of AP is given by<br>T<sub>n&nbsp;<\/sub>= a + (n-1) d<br>T<sub>n<\/sub>= 5 + (n-1) x 6 = 5+ 6n \u2013 6 = 6n \u2013 1<br>(iv) The given AP is (5a \u2013 x), 6a, (7a + x) \u2026..<br>Its first term = (5a \u2013 x) and common difference = 6a \u2013 5a \u2013 x = a + x<br>The n<sup>th<\/sup>&nbsp;term of AP is given by<br>T<sub>n&nbsp;<\/sub>= a + (n-1) d<br>T<sub>11&nbsp;<\/sub>= (5a \u2013 x) + (11-1) (a + x)<br>= 5a \u2013 x + 10x + 10x<br>= 15a + 9x = 3(5a +3x)<br>Hence the 11<sup>th<\/sup>&nbsp;term is 3(5a + 3x)<\/p>\n<p><strong>Question 4:<\/strong><br>(i) The given AP is 63, 58, 53, 48 \u2026.<br>First term = 63, common difference = 58 \u2013 63 = \u2013 5<br>\u2234&nbsp;a = 63, d = \u2013 5<br>The n<sup>th<\/sup>&nbsp;term of AP is given by<br>T<sub>n&nbsp;<\/sub>= a + (n-1) d<br>T<sub>10<\/sub>&nbsp;= 63 + (10-1) (-5) = 63- 45 = 18<br>Hence the 10<sup>th<\/sup>&nbsp;term is 18<br>(ii) The given AP is 9, 5, 1, -3\u2026.<br>First term = 9, common difference = 5 \u2013 9 = -4<br>\u2234&nbsp;a = 9, d= \u2013 4<br>The n<sup>th<\/sup>&nbsp;term of AP is given by<br>T<sub>n&nbsp;<\/sub>= a + (n-1) d<br>T<sub>14<\/sub>&nbsp;= 9 + (14-1) (-4) = 9- 52 = -43<br>Hence, the 14<sup>th<\/sup>&nbsp;term is \u2013 43<br>(iii) The given AP is 16, 9, 2, -5<br>First term = 16, common difference = 9 \u2013 16 = \u2013 7<br>\u2234&nbsp;a = 16, d = -7<br>The n<sup>th<\/sup>&nbsp;term of AP is given by<br>T<sub>n&nbsp;<\/sub>= a + (n-1) d<br>T<sub>n<\/sub>&nbsp;= 16 + (n-1) (-7) \u21d2&nbsp;16- 7n + 7 = (23 \u2013 7n)<br>Hence, the n<sup>th&nbsp;<\/sup>term is (23 \u2013 7n).<br><strong>Question 5:<\/strong><br>The given AP is &nbsp;&nbsp;<span id=\"MathJax-Element-1-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-1\" class=\"math\"><span id=\"MathJax-Span-2\" class=\"mrow\"><span id=\"MathJax-Span-3\" class=\"mn\">6<\/span><span id=\"MathJax-Span-4\" class=\"mo\">,<\/span><span id=\"MathJax-Span-5\" class=\"mn\">7<\/span><span id=\"MathJax-Span-6\" class=\"mfrac\"><span id=\"MathJax-Span-7\" class=\"mn\">3<\/span><span id=\"MathJax-Span-8\" class=\"mn\">4<\/span><\/span><span id=\"MathJax-Span-9\" class=\"mo\">,<\/span><span id=\"MathJax-Span-10\" class=\"mn\">9<\/span><span id=\"MathJax-Span-11\" class=\"mfrac\"><span id=\"MathJax-Span-12\" class=\"mn\">1<\/span><span id=\"MathJax-Span-13\" class=\"mn\">2<\/span><\/span><span id=\"MathJax-Span-14\" class=\"mo\">,<\/span><span id=\"MathJax-Span-15\" class=\"mn\">11<\/span><span id=\"MathJax-Span-16\" class=\"mfrac\"><span id=\"MathJax-Span-17\" class=\"mn\">1<\/span><span id=\"MathJax-Span-18\" class=\"mn\">4<\/span><\/span><span id=\"MathJax-Span-19\" class=\"mo\">\u2026<\/span><span id=\"MathJax-Span-20\" class=\"mo\">\u2026<\/span><span id=\"MathJax-Span-21\" class=\"mo\">.<\/span><span id=\"MathJax-Span-22\" class=\"mo\">.<\/span><\/span><\/span><\/span><br>First term = 6, common difference = &nbsp;&nbsp;<span id=\"MathJax-Element-2-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-23\" class=\"math\"><span id=\"MathJax-Span-24\" class=\"mrow\"><span id=\"MathJax-Span-25\" class=\"mrow\"><span id=\"MathJax-Span-26\" class=\"mo\">(<\/span><span id=\"MathJax-Span-27\" class=\"mn\">7<\/span><span id=\"MathJax-Span-28\" class=\"mfrac\"><span id=\"MathJax-Span-29\" class=\"mn\">3<\/span><span id=\"MathJax-Span-30\" class=\"mn\">4<\/span><\/span><span id=\"MathJax-Span-31\" class=\"mo\">\u2212<\/span><span id=\"MathJax-Span-32\" class=\"mn\">6<\/span><span id=\"MathJax-Span-33\" class=\"mo\">)<\/span><\/span><\/span><\/span><\/span><br>= &nbsp;&nbsp;<span id=\"MathJax-Element-3-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-34\" class=\"math\"><span id=\"MathJax-Span-35\" class=\"mrow\"><span id=\"MathJax-Span-36\" class=\"mrow\"><span id=\"MathJax-Span-37\" class=\"mo\">(<\/span><span id=\"MathJax-Span-38\" class=\"mfrac\"><span id=\"MathJax-Span-39\" class=\"mn\">31<\/span><span id=\"MathJax-Span-40\" class=\"mn\">4<\/span><\/span><span id=\"MathJax-Span-41\" class=\"mo\">\u2212<\/span><span id=\"MathJax-Span-42\" class=\"mn\">6<\/span><span id=\"MathJax-Span-43\" class=\"mo\">)<\/span><\/span><\/span><\/span><\/span><br>= &nbsp;&nbsp;<span id=\"MathJax-Element-4-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-44\" class=\"math\"><span id=\"MathJax-Span-45\" class=\"mrow\"><span id=\"MathJax-Span-46\" class=\"mfrac\"><span id=\"MathJax-Span-47\" class=\"mn\">7<\/span><span id=\"MathJax-Span-48\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span><br>a = 6, d = &nbsp;&nbsp;<span id=\"MathJax-Element-5-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-49\" class=\"math\"><span id=\"MathJax-Span-50\" class=\"mrow\"><span id=\"MathJax-Span-51\" class=\"mfrac\"><span id=\"MathJax-Span-52\" class=\"mn\">7<\/span><span id=\"MathJax-Span-53\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span><br>The n<sup>th<\/sup>&nbsp;term is given by<br>T<sub>n&nbsp;<\/sub>= a + (n-1) d<br>T<sub>14<\/sub>&nbsp;= 6 + (37 \u2013 1) &nbsp;<span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-54\" class=\"math\"><span id=\"MathJax-Span-55\" class=\"mrow\"><span id=\"MathJax-Span-56\" class=\"mo\">(<\/span><span id=\"MathJax-Span-57\" class=\"mfrac\"><span id=\"MathJax-Span-58\" class=\"mn\">7<\/span><span id=\"MathJax-Span-59\" class=\"mn\">4<\/span><\/span><span id=\"MathJax-Span-60\" class=\"mo\">)<\/span><\/span><\/span><\/span>&nbsp;= 6+ 63 = 69<br>Hence, the 37<sup>th<\/sup>&nbsp;term is 69<br><strong>Question 6:<\/strong><br>The given AP is &nbsp; &nbsp;&nbsp;<span id=\"MathJax-Element-7-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-61\" class=\"math\"><span id=\"MathJax-Span-62\" class=\"mrow\"><span id=\"MathJax-Span-63\" class=\"mn\">5<\/span><span id=\"MathJax-Span-64\" class=\"mo\">,<\/span><span id=\"MathJax-Span-65\" class=\"mn\">4<\/span><span id=\"MathJax-Span-66\" class=\"mfrac\"><span id=\"MathJax-Span-67\" class=\"mn\">1<\/span><span id=\"MathJax-Span-68\" class=\"mn\">2<\/span><\/span><span id=\"MathJax-Span-69\" class=\"mo\">,<\/span><span id=\"MathJax-Span-70\" class=\"mn\">4<\/span><span id=\"MathJax-Span-71\" class=\"mo\">,<\/span><span id=\"MathJax-Span-72\" class=\"mn\">3<\/span><span id=\"MathJax-Span-73\" class=\"mfrac\"><span id=\"MathJax-Span-74\" class=\"mn\">1<\/span><span id=\"MathJax-Span-75\" class=\"mn\">2<\/span><\/span><span id=\"MathJax-Span-76\" class=\"mo\">,<\/span><span id=\"MathJax-Span-77\" class=\"mn\">3<\/span><span id=\"MathJax-Span-78\" class=\"mo\">\u2026<\/span><span id=\"MathJax-Span-79\" class=\"mo\">\u2026<\/span><span id=\"MathJax-Span-80\" class=\"mo\">.<\/span><span id=\"MathJax-Span-81\" class=\"mo\">.<\/span><\/span><\/span><\/span><br>The first term = 5,<br>common difference =&nbsp;<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-82\" class=\"math\"><span id=\"MathJax-Span-83\" class=\"mrow\"><span id=\"MathJax-Span-84\" class=\"mrow\"><span id=\"MathJax-Span-85\" class=\"mo\">(<\/span><span id=\"MathJax-Span-86\" class=\"mn\">4<\/span><span id=\"MathJax-Span-87\" class=\"mfrac\"><span id=\"MathJax-Span-88\" class=\"mn\">1<\/span><span id=\"MathJax-Span-89\" class=\"mn\">2<\/span><\/span><span id=\"MathJax-Span-90\" class=\"mo\">\u2212<\/span><span id=\"MathJax-Span-91\" class=\"mn\">5<\/span><span id=\"MathJax-Span-92\" class=\"mo\">)<\/span><\/span><span id=\"MathJax-Span-93\" class=\"mo\">=<\/span><span id=\"MathJax-Span-94\" class=\"mrow\"><span id=\"MathJax-Span-95\" class=\"mo\">(<\/span><span id=\"MathJax-Span-96\" class=\"mfrac\"><span id=\"MathJax-Span-97\" class=\"mn\">9<\/span><span id=\"MathJax-Span-98\" class=\"mn\">2<\/span><\/span><span id=\"MathJax-Span-99\" class=\"mo\">\u2212<\/span><span id=\"MathJax-Span-100\" class=\"mn\">5<\/span><span id=\"MathJax-Span-101\" class=\"mo\">)<\/span><\/span><span id=\"MathJax-Span-102\" class=\"mo\">=<\/span><span id=\"MathJax-Span-103\" class=\"mo\">\u2212<\/span><span id=\"MathJax-Span-104\" class=\"mfrac\"><span id=\"MathJax-Span-105\" class=\"mn\">1<\/span><span id=\"MathJax-Span-106\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span><br>\u2234 a = 5, d =&nbsp;<span id=\"MathJax-Element-9-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-107\" class=\"math\"><span id=\"MathJax-Span-108\" class=\"mrow\"><span id=\"MathJax-Span-109\" class=\"mo\">\u2212<\/span><span id=\"MathJax-Span-110\" class=\"mfrac\"><span id=\"MathJax-Span-111\" class=\"mn\">1<\/span><span id=\"MathJax-Span-112\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span><br>The n<sup>th<\/sup>&nbsp;term is given by<br>T<sub>n&nbsp;<\/sub>= a + (n-1) d<br>T<sub>14<\/sub>&nbsp;= 5 + (25 \u2013 1) (-1\/2) = 5- 12 = -7<br>Hence the 25<sup>th<\/sup>&nbsp;term is \u2013 7<br><strong>Question 7:<\/strong><br>In the given AP, we have a = 6 and d = (10 \u2013 6) = 4<br>Suppose there are n terms in the given AP, then<br>T<sub>n&nbsp;<\/sub>&nbsp;= 174&nbsp;\u21d2 a + (n-1) d = 174<br>\u21d2 6 + (n-1) 4 = 174<br>\u21d2 6 + 4n \u2013 4 = 174<br>\u21d2 2 + 4n = 174&nbsp;\u21d2 &nbsp;n = 172\/4 &nbsp;\u21d2 43<br>Hence there are 43 terms in the given AP<br><strong>Question 8:<\/strong><br>In the given AP we have a = 41 and d = 38 \u2013 41 = \u2013 3<br>Suppose there are n terms in AP, then<br>T<sub>n&nbsp;<\/sub>&nbsp;= 8 \u21d2 a + (n-1) d = 8<br>\u21d2 41 + (n-1) (-3) = 8<br>\u21d2 41 \u2013 3n + 3 = 8<br>\u21d2 -3n = \u2013 36&nbsp;\u21d2 &nbsp;n = 12<br>Hence there are 12 terms in the given AP<br><strong>Question 9:<\/strong><br>In the given AP, we have a = 3 and d = 8 \u2013 3 = 5<br>Suppose there are n terms in a given AP, then<br>T<sub>n&nbsp;<\/sub>&nbsp;= &nbsp;a + (n-1) d = 88<br>\u21d2 3 + (n-1) 5 = 88<br>\u21d2 3 + 5n \u2013 5 = 88<br>\u21d2 5n = 90<br>\u21d2 &nbsp;n = 12<br>Hence, the 18<sup>th<\/sup> term of the given AP is 88<br><strong>Question 10:<\/strong><br>In the given AP, we have a = 72 and d = 68 \u2013 72 = \u2013 4<br>Suppose there are n terms in a given AP, we have<br>T<sub>n&nbsp;<\/sub>&nbsp;= 0 \u21d2 a + (n-1) d =&nbsp;0<br>\u21d2 72 + (n-1) (-4) = 0<br>\u21d2 72 \u2013 4n + 4 = 0<br>\u21d2 4n = 76<br>\u21d2 n = 19<br>Hence, the 19<sup>th<\/sup>&nbsp;term in the given AP is 0<br><strong>Question 11:<\/strong><br>In the given AP, we have &nbsp;a =&nbsp;<span id=\"MathJax-Element-10-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-113\" class=\"math\"><span id=\"MathJax-Span-114\" class=\"mrow\"><span id=\"MathJax-Span-115\" class=\"mfrac\"><span id=\"MathJax-Span-116\" class=\"mn\">1<\/span><span id=\"MathJax-Span-117\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;;&nbsp;<span id=\"MathJax-Element-11-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-118\" class=\"math\"><span id=\"MathJax-Span-119\" class=\"mrow\"><span id=\"MathJax-Span-120\" class=\"mrow\"><span id=\"MathJax-Span-121\" class=\"mo\">(<\/span><span id=\"MathJax-Span-122\" class=\"mn\">1<\/span><span id=\"MathJax-Span-123\" class=\"mo\">\u2212<\/span><span id=\"MathJax-Span-124\" class=\"mfrac\"><span id=\"MathJax-Span-125\" class=\"mn\">5<\/span><span id=\"MathJax-Span-126\" class=\"mn\">6<\/span><\/span><span id=\"MathJax-Span-127\" class=\"mo\">)<\/span><\/span><span id=\"MathJax-Span-128\" class=\"mo\">=<\/span><span id=\"MathJax-Span-129\" class=\"mfrac\"><span id=\"MathJax-Span-130\" class=\"mn\">1<\/span><span id=\"MathJax-Span-131\" class=\"mn\">6<\/span><\/span><\/span><\/span><\/span><br>Suppose there are n terms in a given AP, we have<br>Then,<br>T<sub>n&nbsp;<\/sub>&nbsp;= 3 \u21d2 a + (n-1) d = 3<br>\u21d2 &nbsp;<span id=\"MathJax-Element-12-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-132\" class=\"math\"><span id=\"MathJax-Span-133\" class=\"mrow\"><span id=\"MathJax-Span-134\" class=\"mfrac\"><span id=\"MathJax-Span-135\" class=\"mn\">5<\/span><span id=\"MathJax-Span-136\" class=\"mn\">6<\/span><\/span><span id=\"MathJax-Span-137\" class=\"mo\">+<\/span><span id=\"MathJax-Span-138\" class=\"mo\">(<\/span><span id=\"MathJax-Span-139\" class=\"mi\">n<\/span><span id=\"MathJax-Span-140\" class=\"mo\">\u2212<\/span><span id=\"MathJax-Span-141\" class=\"mn\">1<\/span><span id=\"MathJax-Span-142\" class=\"mo\">)<\/span><span id=\"MathJax-Span-143\" class=\"mfrac\"><span id=\"MathJax-Span-144\" class=\"mn\">1<\/span><span id=\"MathJax-Span-145\" class=\"mn\">6<\/span><\/span><span id=\"MathJax-Span-146\" class=\"mo\">=<\/span><span id=\"MathJax-Span-147\" class=\"mn\">3<\/span><\/span><\/span><\/span><br>\u21d2 &nbsp;<span id=\"MathJax-Element-13-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-148\" class=\"math\"><span id=\"MathJax-Span-149\" class=\"mrow\"><span id=\"MathJax-Span-150\" class=\"mfrac\"><span id=\"MathJax-Span-151\" class=\"mn\">5<\/span><span id=\"MathJax-Span-152\" class=\"mn\">6<\/span><\/span><span id=\"MathJax-Span-153\" class=\"mo\">+<\/span><span id=\"MathJax-Span-154\" class=\"mfrac\"><span id=\"MathJax-Span-155\" class=\"mn\">1<\/span><span id=\"MathJax-Span-156\" class=\"mn\">6<\/span><\/span><span id=\"MathJax-Span-157\" class=\"mi\">n<\/span><span id=\"MathJax-Span-158\" class=\"mo\">\u2212<\/span><span id=\"MathJax-Span-159\" class=\"mfrac\"><span id=\"MathJax-Span-160\" class=\"mn\">1<\/span><span id=\"MathJax-Span-161\" class=\"mn\">6<\/span><\/span><span id=\"MathJax-Span-162\" class=\"mo\">=<\/span><span id=\"MathJax-Span-163\" class=\"mn\">3<\/span><\/span><\/span><\/span><br>\u21d2 4 + n = 18<br>\u21d2 n = 14<br>Thus, 14<sup>th<\/sup>&nbsp;term in the given AP is 3<br><strong>Question 12:<\/strong><br>We know that &nbsp;&nbsp;T<sub>1&nbsp;<\/sub>\u2013 (5x + 2),&nbsp;T<sub>2&nbsp;<\/sub>\u2013 (4x \u2013 1) and&nbsp;&nbsp;T<sub>3&nbsp;<\/sub>\u2013 (x + 2)<br>Clearly,<br>T<sub>2<\/sub>&nbsp;\u2013 T<sub>1<\/sub>&nbsp;= T<sub>3<\/sub>&nbsp;\u2013 T<sub>2<\/sub><br>\u21d2 &nbsp;(4x \u2013 1) \u2013&nbsp;(5x + 2) =&nbsp;(x + 2) \u2013&nbsp;(4x \u2013 1)<br>\u21d2 &nbsp;4x \u2013 1 \u2013 5x \u2013 2 = x + 2 \u2013 4x + 1<br>\u21d2 &nbsp;-x \u2013 3 = -3x + 3<br>\u21d2 &nbsp;-x + 3x = 6<br>\u21d2 &nbsp;2x = 6&nbsp;\u21d2 &nbsp;x = 3<br>Hence x = 3<br><strong>Question 13:<\/strong><br>T<sub>n&nbsp;<\/sub>&nbsp;= (4n \u2013 10)<br>\u21d2&nbsp;T<sub>1&nbsp;<\/sub>&nbsp;= (4 x 1 \u2013 10) = -6 &nbsp;and &nbsp;T<sub>2&nbsp;<\/sub>&nbsp;= (4 x 2 \u2013 10) = -2<br>Thus, we have<br>(i) First term = -6<br>(ii) Common difference &nbsp;=&nbsp;(T<sub>2<\/sub>&nbsp;\u2013 T<sub>1<\/sub>)&nbsp;= (-2+6) = 4<br>(iii) 16<sup>th<\/sup>&nbsp;term = a + (16-1) d, where a = -6 and d = 4<br>= (-6 + 15 x 4) = 54<br><strong>Question 14:<\/strong><br>In the given AP, let the first term = a and common difference = d,<br>Then,&nbsp;T<sub>n&nbsp;<\/sub>&nbsp;= &nbsp;a + (n-1) d<br>\u21d2 T<sub>4&nbsp;<\/sub>&nbsp;= a + (4 \u2013 1)d,&nbsp;T<sub>10&nbsp;<\/sub>&nbsp;= a + (10 \u2013 1)d<br>\u21d2 T<sub>4&nbsp;<\/sub>&nbsp;= a + 3d,&nbsp;T<sub>10&nbsp;<\/sub>&nbsp;= a + 9d<br>Now,&nbsp;T<sub>4&nbsp;<\/sub>&nbsp;= 13 \u21d2 a + 3d = 13 &nbsp;\u2013 \u2013 \u2013 (1)<br>T<sub>10&nbsp;<\/sub>&nbsp;= 25 \u21d2 a + 9d = 25 &nbsp;\u2013 \u2013 \u2013 (2)<br>Subtracting (1) from (2), we get<br>\u21d2 6d = 12&nbsp;\u21d2 d = 2<br>Putting d = 2 in (1), we get<br>a + 3 x 2 = 13<br>\u21d2 a = (13 \u2013 6) = 7<br>Thus, a = 7, and d = 2<br>17<sup>th<\/sup>&nbsp;term = a + (17 \u2013 1)d, where a= 7, d = 2<br>(7 + 16 x 2) = (7 + 32) = 39<br>\u2234&nbsp;a = 7, d = 2,<br><strong>Question 15:<\/strong><br>In the given AP, let the first term = a and common difference = d<br>Then,&nbsp;T<sub>n&nbsp;<\/sub>&nbsp;= &nbsp;a + (n-1) d<br>\u21d2 T<sub>8&nbsp;<\/sub>&nbsp;= a + (8 \u2013 1)d,&nbsp;T<sub>12&nbsp;<\/sub>&nbsp;= a + (12 \u2013 1)d<br>\u21d2 T<sub>8&nbsp;<\/sub>&nbsp;= a + 7d,&nbsp;T<sub>12&nbsp;<\/sub>&nbsp;= a + 11d<br>Now,&nbsp;T<sub>8&nbsp;<\/sub>&nbsp;= 37 \u21d2 a + 7d = 37 &nbsp;\u2013 \u2013 \u2013 (1)<br>T<sub>12&nbsp;<\/sub>&nbsp;= 57 \u21d2 a + 11d = 57 &nbsp;\u2013 \u2013 \u2013 (2)<br>Subtracting (1) from (2), we get<br>\u21d2 4d = 20 \u21d2 d = 5<br>Putting d = 5 in (1), we get<br>a + 7 x 5 = 37<br>\u21d2 a = 2<br>Thus, a = 2, and d = 5<br>So the required AP is 2, 7, 12..<br><strong>Question 16:<\/strong><br>In the given AP, let the first term = a, and common difference = d<br>Then,&nbsp;T<sub>n&nbsp;<\/sub>&nbsp;= &nbsp;a + (n-1) d<br>\u21d2 T<sub>7&nbsp;<\/sub>&nbsp;= a + (7 \u2013 1)d, and T<sub>13&nbsp;<\/sub>&nbsp;= a + (13 \u2013 1)d<br>\u21d2 T<sub>7&nbsp;<\/sub>&nbsp;= a + 6d,&nbsp;T<sub>13&nbsp;<\/sub>&nbsp;= a + 12d<br>Now,&nbsp;T<sub>7&nbsp;<\/sub>&nbsp;= -4 \u21d2 a + 6d = -4 &nbsp;\u2013 \u2013 \u2013 (1)<br>T<sub>13&nbsp;<\/sub>&nbsp;= -16 \u21d2 a + 12d = -16 \u2013 \u2013 \u2013 (2)<br>Subtracting (1) from (2), we get<br>\u21d2 6d = -12 \u21d2 d = -2<br>Putting d = -2 in (1), we get<br>a + 6 &nbsp;(-2) = -4<br>\u21d2 a \u2013 12 = -4<br>\u21d2 a = 8<br>Thus, a = 8, and d = -2<br>So the required AP is 8, 6, 4, 2, 0\u2026\u2026<br><strong>Question 17:<\/strong><br>In the given AP let the first term = a,&nbsp;And common difference = d<br>Then,&nbsp;T<sub>n&nbsp;<\/sub>&nbsp;= &nbsp;a + (n-1) d<br>\u21d2 T<sub>10&nbsp;<\/sub>&nbsp;= a + (10 \u2013 1)d,&nbsp;T<sub>17&nbsp;<\/sub>&nbsp;= a + (17 \u2013 1)d,&nbsp;T<sub>13&nbsp;<\/sub>&nbsp;= a + (13 \u2013 1)d<br>\u21d2 T<sub>10&nbsp;<\/sub>&nbsp;= a + 9d,&nbsp;T<sub>17&nbsp;<\/sub>&nbsp;= a + 16d,&nbsp;T<sub>13&nbsp;<\/sub>&nbsp;= a + 12d<br>Now,&nbsp;T<sub>10&nbsp;<\/sub>&nbsp;= 52 \u21d2 a + 9d = 52 &nbsp;\u2013 \u2013 \u2013 (1)<br>and T<sub>17&nbsp;<\/sub>&nbsp;= T<sub>13&nbsp;<\/sub>+ 20&nbsp;\u21d2 a + 16d = a + 12d + 20<br>\u21d2 4d = 20&nbsp;\u21d2 d = 5<br>Putting d = 5 in (1), we get<br>a + 9 x 5 = 52&nbsp;\u21d2 a = 52-45&nbsp;\u21d2 a = 7<br>Thus, a = 7 and d = 5<br>So the required AP is 7, 12, 17, 22\u2026.<br><strong>Question 18:<\/strong><br>Let the first term of given AP = a and common difference = d<br>Then,&nbsp;T<sub>n&nbsp;<\/sub>&nbsp;= &nbsp;a + (n-1) d<br>\u21d2 T<sub>4&nbsp;<\/sub>&nbsp;= a + (4 \u2013 1)d,&nbsp;T<sub>25&nbsp;<\/sub>&nbsp;= a + (25 \u2013 1)d,&nbsp;T<sub>11&nbsp;<\/sub>&nbsp;= a + (11 \u2013 1)d<br>\u21d2 T<sub>4&nbsp;<\/sub>&nbsp;= a + 3d,&nbsp;T<sub>25&nbsp;<\/sub>&nbsp;= a + 24d,&nbsp;T<sub>11&nbsp;<\/sub>&nbsp;= a + 10d<br>Now,&nbsp;T<sub>4&nbsp;<\/sub>&nbsp;= 0 \u21d2 a + 3d = 0 &nbsp;\u21d2 a = -3d<br>\u2234&nbsp;&nbsp;T<sub>25&nbsp;<\/sub>&nbsp;= a + 24d = (-3d +24d) \u21d2 21d<br>and&nbsp;T<sub>11&nbsp;<\/sub>&nbsp;= a + 10d = (-3d +10d) \u21d2 7d<br>\u2234&nbsp;&nbsp;&nbsp;T<sub>25&nbsp;<\/sub>&nbsp;= 21d = 3 x 7d = 3 x&nbsp;T<sub>11<\/sub><br>Hence 25<sup>th<\/sup>&nbsp;term is triple its 11<sup>th<\/sup>&nbsp;term<br><strong>Question 19:<\/strong><br>The given AP is 3, 8, 13, 18\u2026..<br>First term a = 3, common difference a = 8 \u2013 3 = 5<br>\u2234&nbsp;&nbsp;T<sub>n&nbsp;<\/sub>&nbsp;= &nbsp;a + (n-1) d = 3 + (n \u2013 1) x 5 = 5n \u2013 2<br>T<sub>20&nbsp;<\/sub>&nbsp;= &nbsp;3 + (20-1) 5 = 3 + 19 x 5 = 98<br>Let the n<sup>th<\/sup>&nbsp; term is 55 more than the 20<sup>th<\/sup>&nbsp;term<br>\u2234&nbsp;&nbsp;(5n \u2013 2) \u2013 98 = 55<br>Or 5n = 100 + 55 = 155<br>n = 155\/5 = 31<br>\u2234&nbsp;&nbsp;31<sup>st<\/sup>&nbsp;term is 55 more than the 20<sup>th<\/sup> term of the given AP<br><strong>Question 20:<\/strong><br>The given AP is 5, 15, 25\u2026.<br>a = 5, d = 15 \u2013 5 = 10<br>We have, &nbsp;T<sub>n&nbsp;<\/sub>&nbsp;= &nbsp;130+T<sub>31<\/sub><br>\u21d2 a + (n-1) d = 130 + 5 + (31 \u2013 1) x 10<br>\u21d2 5&nbsp;+ (n-1) 10 = 130 + 5 + (31 \u2013 1) x 10<br>\u21d2 5 + 10n \u2013 10 = 135 + 300<br>\u21d2 10n \u2013 5 = 435 or 10n = 453 + 5<br>\u2234 n = 440\/10 = 44<br>Thus, the required term is 44<sup>th<\/sup><br><strong>Question 21:<\/strong><br>First AP is 63, 65, 67\u2026.<br>First term = 63, common difference = 65 \u2013 63 = 2<br>\u2234 nth term = 63 + (n \u2013 1) 2 = 63 + 2n \u2013 2 = 2n + 61<br>The second AP is 3, 10, 17 \u2026.<br>First term = 3, common difference = 10 \u2013 3 = 7<br>nth term = 3 + (n \u2013 1) 7 = 3 + 7n \u2013 7 = 7n \u2013 4<br>The two nth terms are equal<br>\u2234 2n + 61 = 7n \u2013 4 or 5n = 61 + 4 = 65<br>\u21d2 n = 65\/4 = 13.<br><strong>Question 22:<\/strong><br>Three-digit numbers that are divisible by 7 are 105, 112, 119,\u2026.994<br>This is an AP where a= 105, d = 7 and l = 994<br>Let the n<sup>th<\/sup>&nbsp;term be 994<br>\u2234 a + (n \u2013 1)d =994 or 105 + (n \u2013 1)7 = 994<br>\u21d2 105 + 7n \u2013 7 = 994 or 7n = 94 \u2013 98 = 896<br>\u2234 n = 896\/7 = 128.<br>Hence, there are 128 three digits numbers that are divisible by 7.<br><strong>Question 23:<\/strong><br>Here a = 7, d = (10 \u2013 7) = 3, l = 184<br>And n = 8<br>Now, nth term from the end = [ l \u2013 (n-1) d ]<br>= [ 184 \u2013 (8-1) 3 ]<br>= [ 184 \u2013 7 x 3]<br>= 184-21<br>= 163<br>Hence, the 8<sup>th<\/sup>&nbsp;term from the end is 163<br><strong>Question 24:<\/strong><br>Here a = 17, d = (14 \u2013 17) = -3, l = -40<br>And n = 6<br>Now, an n<sup>th<\/sup>&nbsp;term from the end = [ l \u2013 (n \u2013 1) d ]<br>= [ -40 \u2013 (6-1)(-3) ]<br>= [ -40 + 5 x 3]<br>= -40+15<br>= -25<br>Hence, the 6<sup>th<\/sup>&nbsp;term from the end is \u2013 25<br><strong>Question 25:<\/strong><br>The given AP is 10, 7, 4, \u2026.. (-62)<br>a = 10, d = 7 \u2013 10 = -3, l = -62<br>Now, the 11<sup>th<\/sup>&nbsp;term from the end = [ l \u2013 (n \u2013 1) d ]<br>= [ -62 \u2013 (11-1)(-3) ]<br>= -62 + 30<br>= -32<br><strong>Question 26:<\/strong><br>Let a be the first term and d be the common difference<br>p<sup>th<\/sup>&nbsp;term = a +(p \u2013 1)d = q (given) \u2014\u2013(1)<br>q<sup>th<\/sup>&nbsp;term = a +(q \u2013 1) d = p (given) \u2014\u2013(2)<br>subtracting (2) from (1)<br>(p \u2013 q)d = q \u2013 p<br>(p \u2013 q)d = -(p \u2013 q)<br>d = -1<br>Putting d = -1 in (1)<br>a \u2013 (p \u2013 1) = q &nbsp;&nbsp;\u2234 a = p + q -1<br>\u2234 (p + q)th term = a+ (p + q -1)d<br>=&nbsp;(p + q -1) \u2013&nbsp;(p + q -1) = 0<br><strong>Question 27:<\/strong><br>Let a be the first term and d be the common difference<br>T<sub>10&nbsp;<\/sub>&nbsp;= a + 9d, &nbsp;T<sub>15&nbsp;<\/sub>&nbsp;= &nbsp;a + 14d<br>10T<sub>10<\/sub>&nbsp;=&nbsp;15T<sub>15<\/sub><br>\u21d2 10(a + 9) d = 15(a + 14d)<br>\u21d2 2(a + 9) d = 3(a + 14d)<br>\u21d2 a + 24d = 0<br>\u2234 T<sub>25&nbsp;<\/sub>&nbsp;= 0<br><strong>Question 28:<\/strong><br>Let a be the first term and d be the common difference<br>\u2234 &nbsp;n<sup>th<\/sup>&nbsp;term from the beginning = a + (n \u2013 1)d \u2014\u2013(1)<br>n<sup>th<\/sup>&nbsp;term from end = l \u2013 (n \u2013 1)d \u2014-(2)<br>adding (1) and (2),<br>the sum of the n<sup>th<\/sup> term from the beginning and the n<sup>th<\/sup>&nbsp;term from the end = [a + (n \u2013 1)d] + [l \u2013 (n \u2013 1)d] = a + l<br><strong>Question 29:<\/strong><br>The number of rose plants in the first, second, and third rows\u2026. is 43, 41, and 39\u2026 respectively.<br>There are 11 rose plants in the last row<br>So, it is an AP. viz. 43, 41, 39 \u2026. 11<br>a = 43, d = 41 \u2013 43 = -2, l = 11<br>Let the n<sup>th<\/sup>&nbsp;term be the last term<br>\u2234 l<sub>&nbsp;<\/sub>&nbsp;= a + (n-1) d<br>\u21d2 11 = 43 + (n-1) x (-2)<br>43 \u2013 2n + 2 = 11 or 2n = 45 -11 = 34<br>\u2234 n = 34\/2 = 17<br>Hence, there are 17 rows in the flower bed.<br><strong>Question 30:<\/strong><br>Total amount = \u20b9 2800<br>and number of prizes = 4<br>Let first prize = \u20b9 a<br>The second prize = \u20b9 a \u2013 200<br>Third prize = a \u2013 200 \u2013 200 = a \u2013 400<br>and fourth prize = a \u2013 400 \u2013 200 = a \u2013 600<br>But sum of there 4 prizes are \u20b9 2800<br>a + a \u2013 200 + a \u2013 400 + a \u2013 600 = \u20b9 2800<br>\u21d2 4a \u2013 1200 = 2800<br>\u21d2 4a = 2800 + 1200 = 4000<br>\u21d2 a = 1000<br>First prize = \u20b9 1000<br>Second prize = \u20b9 1000 \u2013 200 = \u20b9 800<br>Third prize = \u20b9 800 \u2013 200 = \u20b9 600<br>and fourth prize = \u20b9 600 \u2013 200 = \u20b9 400<\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rs-aggarwal-solutions-class-10-maths-chapter-11-arithmetic-progression-other-important-exercises\"><\/span>Access RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression Other Important Exercises<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-11-class-10-maths-exercise-11-1-solutions\/\" target=\"_blank\" rel=\"noopener\">RS Aggarwal Solutions Chapter 11 Exercise 11.1<\/a><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-11-class-10-maths-exercise-11-2-solutions\/\" target=\"_blank\" rel=\"noopener\">RS Aggarwal Solutions Chapter 11 Exercise 11.2<\/a><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-11-class-10-maths-exercise-11-3-solutions\/\" target=\"_blank\" rel=\"noopener\">RS Aggarwal Solutions Chapter 11 Exercise 11.3<\/a><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-11-class-10-maths-exercise-11-4-solutions\/\" target=\"_blank\" rel=\"noopener\">RS Aggarwal Solutions Chapter 11 Exercise 11.4<\/a><\/p>\n<p>This is the complete blog on the RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression. To know more about the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 10 Maths exam, ask in the comments.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rs-aggarwal-solutions-class-10-maths-chapter-11\"><\/span>FAQs on RS Aggarwal Solutions Class 10 Maths Chapter 11<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-1633449147633\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-find-the-download-link-for-the-rs-aggarwal-solutions-class-10-maths-chapter-11-arithmetic-progression-pdf\"><\/span>From where can I find the download link for the RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link for the RS Aggarwal Solutions Class 10 Maths Chapter 11 PDF in the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1633449164388\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-rs-aggarwal-solutions-class-10-maths-chapter-11-arithmetic-progression-pdf\"><\/span>How much does it cost to download the\u00a0RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RS Aggarwal Solutions Class 10 Maths Chapter 11 PDF for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1633449196402\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rs-aggarwal-solutions-class-10-maths-chapter-11-arithmetic-progression-pdf-offline\"><\/span>Can I access the\u00a0RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression PDF online, you can access it offline whenever you want.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1635098521729\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-exercises-are-there-in-rs-aggarwal-solutions-class-10-maths-chapter-11-arithmetic-progression\"><\/span>How many exercises are there in RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are 4 exercises in RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression: Download the Free PDF of RS Aggarwal Solutions Class 10 Maths Chapter 11 from here. All the solutions of RS Aggarwal Solutions Class 10 Maths are prepared by subject matter experts and are as per the current CBSE Syllabus. 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