{"id":67989,"date":"2023-09-13T19:00:00","date_gmt":"2023-09-13T13:30:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67989"},"modified":"2023-11-15T10:13:30","modified_gmt":"2023-11-15T04:43:30","slug":"rd-sharma-solutions-class-11-maths-chapter-18-exercise-18-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18-exercise-18-2\/","title":{"rendered":"RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-121973\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-18-Exercise-18.2.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-18-Exercise-18.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-18-Exercise-18.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2<\/strong>: An algebraic expression made up of two words is called a binomial expression. For students who want to learn the correct steps to solve such problems, <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18-binomial-theorem\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 18<\/a> Exercise 18.2 is available in PDF format, and students can easily download the PDF from the link provided below.<\/p>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths<\/a>&nbsp;<\/li>\n<\/ul>\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 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id=\"item-69da38a783558\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18-exercise-18-2\/#download-rd-sharma-solutions-class-11-maths-chapter-18-exercise-182-pdf\" title=\"Download RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2 PDF:\">Download RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2 PDF:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18-exercise-18-2\/#access-rd-sharma-solutions-class-11-maths-chapter-18-exercise-182\" title=\"Access RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2\">Access RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18-exercise-18-2\/#faq-rd-sharma-solutions-class-11-maths-chapter-18-exercise-182\" title=\"FAQ: RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2\">FAQ: RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18-exercise-18-2\/#what-are-the-benefits-of-studying-from-rd-sharma-solutions-class-12\" title=\"What are the benefits of studying from RD Sharma Solutions Class 12?\">What are the benefits of studying from RD Sharma Solutions Class 12?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" 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src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-18-Ex-18.2-1.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-18-Ex-18.2-1.pdf\">RD-Sharma-Solutions-Class-11-Maths-Chapter-18-Ex-18.2<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-solutions-class-11-maths-chapter-18-exercise-182\"><\/span>Access RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>1. Find the 11<sup>th<\/sup>&nbsp;term from the beginning and the 11<sup>th<\/sup>&nbsp;term from the end in the expansion of (2x \u2013 1\/x<sup>2<\/sup>)<sup>25<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(2x \u2013 1\/x<sup>2<\/sup>)<sup>25<\/sup><\/p>\n<p>The given expression contains 26 terms.<\/p>\n<p>So, the 11<sup>th<\/sup>&nbsp;term from the end is the (26 \u2212 11 + 1)&nbsp;<sup>th<\/sup>&nbsp;term from the beginning.<\/p>\n<p>In other words, the 11<sup>th<\/sup>&nbsp;term from the end is the 16<sup>th<\/sup>&nbsp;term from the beginning.<\/p>\n<p>Then,<\/p>\n<p>T<sub>16<\/sub>&nbsp;= T<sub>15+1<\/sub>&nbsp;=&nbsp;<sup>25<\/sup>C<sub>15<\/sub>&nbsp;(2x)<sup>25-15<\/sup>&nbsp;(-1\/x<sup>2<\/sup>)<sup>15<\/sup><\/p>\n<p>=&nbsp;<sup>25<\/sup>C<sub>15<\/sub>&nbsp;(2<sup>10<\/sup>) (x)<sup>10<\/sup>&nbsp;(-1\/x<sup>30<\/sup>)<\/p>\n<p>= \u2013&nbsp;<sup>25<\/sup>C<sub>15<\/sub>&nbsp;(2<sup>10<\/sup>&nbsp;\/ x<sup>20<\/sup>)<\/p>\n<p>Now, we shall find the 11<sup>th<\/sup>&nbsp;term from the beginning.<\/p>\n<p>T<sub>11<\/sub>&nbsp;= T<sub>10+1<\/sub>&nbsp;=&nbsp;<sup>25<\/sup>C<sub>10<\/sub>&nbsp;(2x)<sup>25-10<\/sup>&nbsp;(-1\/x<sup>2<\/sup>)<sup>10<\/sup><\/p>\n<p>=&nbsp;<sup>25<\/sup>C<sub>10<\/sub>&nbsp;(2<sup>15<\/sup>) (x)<sup>15<\/sup>&nbsp;(1\/x<sup>20<\/sup>)<\/p>\n<p>=&nbsp;<sup>25<\/sup>C<sub>10<\/sub>&nbsp;(2<sup>15<\/sup>&nbsp;\/ x<sup>5<\/sup>)<\/p>\n<p><strong>2. Find the 7<sup>th<\/sup>&nbsp;term in the expansion of (3x<sup>2<\/sup>&nbsp;\u2013 1\/x<sup>3<\/sup>)<sup>10<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(3x<sup>2<\/sup>&nbsp;\u2013 1\/x<sup>3<\/sup>)<sup>10<\/sup><\/p>\n<p>Let us consider the 7<sup>th<\/sup>&nbsp;term as T<sub>7<\/sub><\/p>\n<p>So,<\/p>\n<p>T<sub>7<\/sub>&nbsp;= T<sub>6+1<\/sub><\/p>\n<p>=&nbsp;<sup>10<\/sup>C<sub>6<\/sub>&nbsp;(3x<sup>2<\/sup>)<sup>10-6<\/sup>&nbsp;(-1\/x<sup>3<\/sup>)<sup>6<\/sup><\/p>\n<p>=&nbsp;<sup>10<\/sup>C<sub>6<\/sub>&nbsp;(3)<sup>4<\/sup>&nbsp;(x)<sup>8<\/sup>&nbsp;(1\/x<sup>18<\/sup>)<\/p>\n<p>= [10\u00d79\u00d78\u00d77\u00d781] \/ [4\u00d73\u00d72\u00d7x<sup>10<\/sup>]<\/p>\n<p>= 17010 \/ x<sup>10<\/sup><\/p>\n<p>\u2234 The 7<sup>th<\/sup>&nbsp;term of the expression (3x<sup>2<\/sup>&nbsp;\u2013 1\/x<sup>3<\/sup>)<sup>10<\/sup>&nbsp;is 17010 \/ x<sup>10<\/sup>.<\/p>\n<p><strong>3.<\/strong>&nbsp;<strong>Find the 5<sup>th<\/sup>&nbsp;term in the expansion of (3x \u2013 1\/x<sup>2<\/sup>)<sup>10<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(3x \u2013 1\/x<sup>2<\/sup>)<sup>10<\/sup><\/p>\n<p>The 5<sup>th<\/sup>&nbsp;term from the end is the (11 \u2013 5 + 1)th, is., 7<sup>th<\/sup>&nbsp;term from the beginning.<\/p>\n<p>So,<\/p>\n<p>T<sub>7<\/sub>&nbsp;= T<sub>6+1<\/sub><\/p>\n<p>=&nbsp;<sup>10<\/sup>C<sub>6<\/sub>&nbsp;(3x)<sup>10-6<\/sup>&nbsp;(-1\/x<sup>2<\/sup>)<sup>6<\/sup><\/p>\n<p>=&nbsp;<sup>10<\/sup>C<sub>6<\/sub>&nbsp;(3)<sup>4<\/sup>&nbsp;(x)<sup>4<\/sup>&nbsp;(1\/x<sup>12<\/sup>)<\/p>\n<p>= [10\u00d79\u00d78\u00d77\u00d781] \/ [4\u00d73\u00d72\u00d7x<sup>8<\/sup>]<\/p>\n<p>= 17010 \/ x<sup>8<\/sup><\/p>\n<p>\u2234 The 5<sup>th<\/sup>&nbsp;term of the expression (3x \u2013 1\/x<sup>2<\/sup>)<sup>10<\/sup>&nbsp;is 17010 \/ x<sup>8<\/sup>.<\/p>\n<p><strong>4. Find the 8<sup>th<\/sup>&nbsp;term in the expansion of (x<sup>3\/2<\/sup>&nbsp;y<sup>1\/2<\/sup>&nbsp;\u2013 x<sup>1\/2<\/sup>&nbsp;y<sup>3\/2<\/sup>)<sup>10<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(x<sup>3\/2<\/sup>&nbsp;y<sup>1\/2<\/sup>&nbsp;\u2013 x<sup>1\/2<\/sup>&nbsp;y<sup>3\/2<\/sup>)<sup>10<\/sup><\/p>\n<p>Let us consider the 8<sup>th<\/sup>&nbsp;term as T<sub>8<\/sub><\/p>\n<p>So,<\/p>\n<p>T<sub>8<\/sub>&nbsp;= T<sub>7+1<\/sub><\/p>\n<p>=&nbsp;<sup>10<\/sup>C<sub>7<\/sub>&nbsp;(x<sup>3\/2<\/sup>&nbsp;y<sup>1\/2<\/sup>)<sup>10-7<\/sup>&nbsp;(-x<sup>1\/2<\/sup>&nbsp;y<sup>3\/2<\/sup>)<sup>7<\/sup><\/p>\n<p>= -[10\u00d79\u00d78]\/[3\u00d72] x<sup>9\/2<\/sup>&nbsp;y<sup>3\/2<\/sup>&nbsp;(x<sup>7\/2<\/sup>&nbsp;y<sup>21\/2<\/sup>)<\/p>\n<p>= -120 x<sup>8<\/sup>y<sup>12<\/sup><\/p>\n<p>\u2234 The 8<sup>th<\/sup>&nbsp;term of the expression (x<sup>3\/2<\/sup>&nbsp;y<sup>1\/2<\/sup>&nbsp;\u2013 x<sup>1\/2<\/sup>&nbsp;y<sup>3\/2<\/sup>)<sup>10<\/sup>&nbsp;is -120 x<sup>8<\/sup>y<sup>12<\/sup>.<\/p>\n<p><strong>5. Find the 7<sup>th<\/sup>&nbsp;term in the expansion of (4x\/5 + 5\/2x)<sup>&nbsp;8<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(4x\/5 + 5\/2x)<sup>&nbsp;8<\/sup><\/p>\n<p>Let us consider the 7<sup>th<\/sup>&nbsp;term as T<sub>7<\/sub><\/p>\n<p>So,<\/p>\n<p>T<sub>7<\/sub>&nbsp;= T<sub>6+1<\/sub><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-21.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 21\"><\/p>\n<p>\u2234 The 7<sup>th<\/sup>&nbsp;term of the expression (4x\/5 + 5\/2x)<sup>&nbsp;8<\/sup>&nbsp;is 4375\/x<sup>4<\/sup>.<\/p>\n<p><strong>6. Find the 4<sup>th<\/sup>&nbsp;term from the beginning and 4<sup>th<\/sup>&nbsp;term from the end in the expansion of (x + 2\/x)<sup>&nbsp;9<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(x + 2\/x)<sup>&nbsp;9<\/sup><\/p>\n<p>Let&nbsp;T<sub>r+1<\/sub>&nbsp;be the 4th term from the end.<\/p>\n<p>Then,&nbsp;T<sub>r+1<\/sub>&nbsp;is (10 \u2212 4 + 1)th, i.e., 7th, the term from the beginning.<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-22.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 22\"><\/p>\n<p><strong>7. Find the 4<sup>th<\/sup>&nbsp;term from the end in the expansion of (4x\/5 \u2013 5\/2x)<sup>&nbsp;9<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(4x\/5 \u2013 5\/2x)<sup>&nbsp;9<\/sup><\/p>\n<p>Let T<sub>r+1<\/sub>&nbsp;be the<sub>&nbsp;<\/sub>4th term from the end of the given expression.<\/p>\n<p>Then, T<sub>r+1&nbsp;<\/sub>is (10 \u2212 4 + 1)th term, i.e., the 7th term, from the beginning.<\/p>\n<p>T<sub>7<\/sub>&nbsp;= T<sub>6+1<\/sub><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-23.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 23\"><\/p>\n<p>\u2234 The 4<sup>th<\/sup>&nbsp;term from the end is 10500\/x<sup>3<\/sup>.<\/p>\n<p><strong>8. Find the 7th term from the end in the expansion of (2x<sup>2<\/sup>&nbsp;\u2013 3\/2x)<sup>&nbsp;8<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(2x<sup>2<\/sup>&nbsp;\u2013 3\/2x)<sup>&nbsp;8<\/sup><\/p>\n<p>Let T<sub>r+1<\/sub>&nbsp;be the<sub>&nbsp;<\/sub>4th term from the end of the given expression.<\/p>\n<p>Then, T<sub>r+1&nbsp;<\/sub>is (9 \u2212 7 + 1)th term, i.e., 3rd term, from the beginning.<\/p>\n<p>T<sub>3<\/sub>&nbsp;= T<sub>2+1<\/sub><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-24.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 24\"><\/p>\n<p>\u2234 The 7<sup>th<\/sup>&nbsp;term from the end is 4032 x<sup>10<\/sup>.<\/p>\n<p><strong>9. Find the coefficient of:<\/strong><\/p>\n<p><strong>(i) &nbsp;x<sup>10<\/sup>&nbsp;in the expansion of (2x<sup>2<\/sup>&nbsp;\u2013 1\/x)<sup>20<\/sup><\/strong><\/p>\n<p><strong>(ii) x<sup>7<\/sup>&nbsp;in the expansion of (x \u2013 1\/x<sup>2<\/sup>)<sup>40<\/sup><\/strong><\/p>\n<p><strong>(iii) x<sup>-15<\/sup>&nbsp;in the expansion of (3x<sup>2<\/sup>&nbsp;\u2013 a\/3x<sup>3<\/sup>)<sup>10<\/sup><\/strong><\/p>\n<p><strong>(iv) x<sup>9<\/sup>&nbsp;in the expansion of (x<sup>2<\/sup>&nbsp;\u2013 1\/3x)<sup>9<\/sup><\/strong><\/p>\n<p><strong>(v) x<sup>m<\/sup>&nbsp;in the expansion of (x + 1\/x)<sup>n<\/sup><\/strong><\/p>\n<p><strong>(vi) x in the expansion of (1 \u2013 2x<sup>3<\/sup>&nbsp;+ 3x<sup>5<\/sup>) (1 + 1\/x)<sup>8<\/sup><\/strong><\/p>\n<p><strong>(vii) a<sup>5<\/sup>b<sup>7<\/sup>&nbsp;in the expansion of (a \u2013 2b)<sup>12<\/sup><\/strong><\/p>\n<p><strong>(viii) x in the expansion of (1 \u2013 3x + 7x<sup>2<\/sup>) (1 \u2013 x)<sup>16<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i) &nbsp;<\/strong>x<sup>10<\/sup>&nbsp;in the expansion of (2x<sup>2<\/sup>&nbsp;\u2013 1\/x)<sup>20<\/sup><\/p>\n<p>Given:<\/p>\n<p>(2x<sup>2<\/sup>&nbsp;\u2013 1\/x)<sup>20<\/sup><\/p>\n<p>If &nbsp;x<sup>10&nbsp;<\/sup>occurs in the (r&nbsp;+ 1)th term in the given expression.<\/p>\n<p>Then, we have:<\/p>\n<p>T<sub>r+1&nbsp;<\/sub>=&nbsp;<sup>n<\/sup>C<sub>r<\/sub>&nbsp;x<sup>n-r<\/sup>&nbsp;a<sup>r<\/sup><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-25.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 25\"><\/p>\n<p><strong>(ii)&nbsp;<\/strong>x<sup>7<\/sup>&nbsp;in the expansion of (x \u2013 1\/x<sup>2<\/sup>)<sup>40<\/sup><\/p>\n<p>Given:<\/p>\n<p>(x \u2013 1\/x<sup>2<\/sup>)<sup>40<\/sup><\/p>\n<p>If x<sup>7&nbsp;<\/sup>occurs at the (r&nbsp;+ 1) th term in the given expression.<\/p>\n<p>Then, we have:<\/p>\n<p>T<sub>r+1&nbsp;<\/sub>=&nbsp;<sup>n<\/sup>C<sub>r<\/sub>&nbsp;x<sup>n-r<\/sup>&nbsp;a<sup>r<\/sup><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-26.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 26\"><\/p>\n<p>40 \u2212 3r&nbsp;=7<\/p>\n<p>3r = 40 \u2013 7<\/p>\n<p>3r = 33<\/p>\n<p>r = 33\/3<\/p>\n<p>= 11<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-27.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 27\"><\/p>\n<p><strong>(iii)&nbsp;<\/strong>x<sup>-15<\/sup>&nbsp;in the expansion of (3x<sup>2<\/sup>&nbsp;\u2013 a\/3x<sup>3<\/sup>)<sup>10<\/sup><\/p>\n<p>Given:<\/p>\n<p>(3x<sup>2<\/sup>&nbsp;\u2013 a\/3x<sup>3<\/sup>)<sup>10<\/sup><\/p>\n<p>If&nbsp;x<sup>\u221215<\/sup>&nbsp;occurs at the (r&nbsp;+ 1)th term in the given expression.<\/p>\n<p>Then, we have:<\/p>\n<p>T<sub>r+1&nbsp;<\/sub>=&nbsp;<sup>n<\/sup>C<sub>r<\/sub>&nbsp;x<sup>n-r<\/sup>&nbsp;a<sup>r<\/sup><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-28.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 28\"><\/p>\n<p><strong>(iv)&nbsp;<\/strong>x<sup>9<\/sup>&nbsp;in the expansion of (x<sup>2<\/sup>&nbsp;\u2013 1\/3x)<sup>9<\/sup><\/p>\n<p>Given:<\/p>\n<p>(x<sup>2<\/sup>&nbsp;\u2013 1\/3x)<sup>9<\/sup><\/p>\n<p>If&nbsp;x<sup>9<\/sup>&nbsp;occurs at the (r&nbsp;+ 1)th term in the above expression.<\/p>\n<p>Then, we have:<\/p>\n<p>T<sub>r+1&nbsp;<\/sub>=&nbsp;<sup>n<\/sup>C<sub>r<\/sub>&nbsp;x<sup>n-r<\/sup>&nbsp;a<sup>r<\/sup><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-29.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 29\"><\/p>\n<p>For&nbsp;this&nbsp;term&nbsp;to&nbsp;contain&nbsp;x<sup>9<\/sup>, we must have<\/p>\n<p>18 \u2212 3r&nbsp;= 9<\/p>\n<p>3r = 18 \u2013 9<\/p>\n<p>3r = 9<\/p>\n<p>r = 9\/3<\/p>\n<p>= 3<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-30.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 30\"><\/p>\n<p><strong>(v)&nbsp;<\/strong>x<sup>m<\/sup>&nbsp;in the expansion of (x + 1\/x)<sup>n<\/sup><\/p>\n<p>Given:<\/p>\n<p>(x + 1\/x)<sup>n<\/sup><\/p>\n<p>If&nbsp;x<sup>m<\/sup>&nbsp;occurs at the (r&nbsp;+ 1)th term in the given expression.<\/p>\n<p>Then, we have:<\/p>\n<p>T<sub>r+1&nbsp;<\/sub>=&nbsp;<sup>n<\/sup>C<sub>r<\/sub>&nbsp;x<sup>n-r<\/sup>&nbsp;a<sup>r<\/sup><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-31.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 31\"><\/p>\n<p><strong>(vi)&nbsp;<\/strong>x in the expansion of (1 \u2013 2x<sup>3<\/sup>&nbsp;+ 3x<sup>5<\/sup>) (1 + 1\/x)<sup>8<\/sup><\/p>\n<p>Given:<\/p>\n<p>(1 \u2013 2x<sup>3<\/sup>&nbsp;+ 3x<sup>5<\/sup>) (1 + 1\/x)<sup>8<\/sup><\/p>\n<p>If&nbsp;x&nbsp;occurs at the (r&nbsp;+ 1)th term in the given expression.<\/p>\n<p>Then, we have:<\/p>\n<p>(1 \u2013 2x<sup>3<\/sup>&nbsp;+ 3x<sup>5<\/sup>) (1 + 1\/x)<sup>8<\/sup><strong>&nbsp;=&nbsp;<\/strong>(1 \u2013 2x<sup>3<\/sup>&nbsp;+ 3x<sup>5<\/sup>) (<sup>8<\/sup>C<sub>0<\/sub>&nbsp;+&nbsp;<sup>8<\/sup>C<sub>1<\/sub>&nbsp;(1\/x) +&nbsp;<sup>8<\/sup>C<sub>2<\/sub>&nbsp;(1\/x)<sup>2<\/sup>&nbsp;+&nbsp;<sup>8<\/sup>C<sub>3<\/sub>&nbsp;(1\/x)<sup>3<\/sup>&nbsp;+&nbsp;<sup>8<\/sup>C<sub>4<\/sub>&nbsp;(1\/x)<sup>4<\/sup>&nbsp;+&nbsp;<sup>8<\/sup>C<sub>5<\/sub>&nbsp;(1\/x)<sup>5<\/sup>&nbsp;+&nbsp;<sup>8<\/sup>C<sub>6<\/sub>&nbsp;(1\/x)<sup>6<\/sup>&nbsp;+&nbsp;<sup>8<\/sup>C<sub>7<\/sub>&nbsp;(1\/x)<sup>7<\/sup>&nbsp;+&nbsp;<sup>8<\/sup>C<sub>8<\/sub>&nbsp;(1\/x)<sup>8<\/sup>)<\/p>\n<p>So, \u2018x\u2019 occurs in the above expression at -2x<sup>3<\/sup>.<sup>8<\/sup>C<sub>2<\/sub>&nbsp;(1\/x<sup>2<\/sup>) + 3x<sup>5<\/sup>.<sup>8<\/sup>C<sub>4<\/sub>&nbsp;(1\/x<sup>4<\/sup>)<\/p>\n<p>\u2234 Coefficient of x = -2 (8!\/(2!6!)) + 3 (8!\/(4! 4!))<\/p>\n<p>= -56 + 210<\/p>\n<p>= 154<\/p>\n<p><strong>(vii)&nbsp;<\/strong>a<sup>5<\/sup>b<sup>7<\/sup>&nbsp;in the expansion of (a \u2013 2b)<sup>12<\/sup><\/p>\n<p>Given:<\/p>\n<p>(a \u2013 2b)<sup>12<\/sup><\/p>\n<p>If&nbsp;a<sup>5<\/sup>b<sup>7<\/sup>&nbsp;occurs at the (r&nbsp;+ 1)th term in the given expression.<\/p>\n<p>Then, we have:<\/p>\n<p>T<sub>r+1&nbsp;<\/sub>=&nbsp;<sup>n<\/sup>C<sub>r<\/sub>&nbsp;x<sup>n-r<\/sup>&nbsp;a<sup>r<\/sup><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-32.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 32\"><\/p>\n<p><strong>(viii)&nbsp;<\/strong>x in the expansion of (1 \u2013 3x + 7x<sup>2<\/sup>) (1 \u2013 x)<sup>16<\/sup><\/p>\n<p>Given:<\/p>\n<p>(1 \u2013 3x + 7x<sup>2<\/sup>) (1 \u2013 x)<sup>16<\/sup><\/p>\n<p>If&nbsp;x&nbsp;occurs at the (r&nbsp;+ 1)th term in the given expression.<\/p>\n<p>Then, we have:<\/p>\n<p>(1 \u2013 3x + 7x<sup>2<\/sup>) (1 \u2013 x)<sup>16<\/sup>&nbsp;= (1 \u2013 3x + 7x<sup>2<\/sup>) (<sup>16<\/sup>C<sub>0<\/sub>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>1<\/sub>&nbsp;(-x) +&nbsp;<sup>16<\/sup>C<sub>2<\/sub>&nbsp;(-x)<sup>2<\/sup>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>3<\/sub>&nbsp;(-x)<sup>3<\/sup>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>4<\/sub>&nbsp;(-x)<sup>4<\/sup>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>5<\/sub>&nbsp;(-x)<sup>5<\/sup>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>6<\/sub>&nbsp;(-x)<sup>6<\/sup>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>7<\/sub>&nbsp;(-x)<sup>7<\/sup>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>8<\/sub>&nbsp;(-x)<sup>8<\/sup>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>9<\/sub>&nbsp;(-x)<sup>9<\/sup>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>10<\/sub>&nbsp;(-x)<sup>10<\/sup>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>11<\/sub>&nbsp;(-x)<sup>11<\/sup>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>12<\/sub>&nbsp;(-x)<sup>12<\/sup>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>13<\/sub>&nbsp;(-x)<sup>13<\/sup>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>14<\/sub>&nbsp;(-x)<sup>14<\/sup>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>15<\/sub>&nbsp;(-x)<sup>15<\/sup>&nbsp;+&nbsp;<sup>16<\/sup>C<sub>16<\/sub>&nbsp;(-x)<sup>16<\/sup>)<\/p>\n<p>So, \u2018x\u2019 occurs in the above expression at&nbsp;<sup>16<\/sup>C<sub>1<\/sub>&nbsp;(-x) \u2013 3x<sup>16<\/sup>C<sub>0<\/sub><\/p>\n<p>\u2234 Coefficient of x = -(16!\/(1! 15!)) \u2013 3(16!\/(0! 16!))<\/p>\n<p>= -16 \u2013 3<\/p>\n<p>= -19<\/p>\n<p><strong>10. Which term in the expansion of&nbsp;<img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-33.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 33\">contains x and y to one and the same power?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider T<sub>r+1&nbsp;<\/sub>th term in the given expansion contains x and y to one and the same power.<\/p>\n<p>Then we have,<\/p>\n<p>T<sub>r+1<\/sub>&nbsp;=&nbsp;<sup>n<\/sup>C<sub>r<\/sub>&nbsp;x<sup>n-r<\/sup>&nbsp;a<sup>r<\/sup><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-34.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 34\"><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-35.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 35\"><\/p>\n<p>We have included all the information regarding&nbsp;<a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2. If you have any queries feel free to ask in the comment section.&nbsp;<\/p>\n<h3><span class=\"ez-toc-section\" id=\"faq-rd-sharma-solutions-class-11-maths-chapter-18-exercise-182\"><\/span>FAQ: RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630416243539\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-studying-from-rd-sharma-solutions-class-12\"><\/span>What are the benefits of studying from RD Sharma Solutions Class 12?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practicing these solutions, students can earn higher academic grades. Our experts solve these solutions with utmost accuracy to help students in their studies.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630416256845\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-download-rd-sharma-solutions-class-11-maths-chapter-18-exercise-182-pdf-free\"><\/span>Can I download RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2 PDF free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can download RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630416264733\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-enough-for-class-12-maths\"><\/span>Is RD Sharma enough for Class 12 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>RD Sharma is a good book that gives you thousands of questions to practice.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630416286989\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-open-rd-sharma-solutions-class-11-maths-chapter-18-exercise-182-pdf-on-my-smartphone\"><\/span>Can I open RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2 PDF on my smartphone?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can open RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2 PDF on any device.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2: An algebraic expression made up of two words is called a binomial expression. For students who want to learn the correct steps to solve such problems, RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2 is available in PDF format, and students can easily &#8230; <a title=\"RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18-exercise-18-2\/\" aria-label=\"More on RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":121973,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2917,73397,73717,73411,2985,73410],"tags":[3428,73334,73564],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67989"}],"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\/244"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=67989"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67989\/revisions"}],"predecessor-version":[{"id":507262,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67989\/revisions\/507262"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/121973"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67989"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67989"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67989"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}