{"id":5257,"date":"2019-05-30T23:02:53","date_gmt":"2019-05-30T17:32:53","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=5257"},"modified":"2022-11-04T12:06:29","modified_gmt":"2022-11-04T06:36:29","slug":"arithmetic-progression-questions-class-10","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/arithmetic-progression-questions-class-10\/","title":{"rendered":"Arithmetic progression questions for class 10"},"content":{"rendered":"
This page contains arithmetic progression questions for class 10. In this link get short arithmetic progression class 10 notes<\/a>. Few other links are\r\narithmetic progression class 10 important questions<\/a>\r\narithmetic progression assignment<\/a>\r\nPracticing these questions on Arithmetic progression for class 10 maths<\/a> can be helpful for your board exams.<\/pre>\n
\n
Arithmetic Progression questions and answers <\/strong><\/h2>\n
class 10 maths<\/span><\/strong><\/h2>\n
\n
One Marks Questions<\/strong><\/span><\/h3>\n
Question 1. <\/strong>If k, 2k – 1 and 2k + 1 are three consecutive terms of an A.P., find the value of k.
\nQuestion 2. <\/strong>Find the common difference of the A.P. \\(\\frac{1}{2b},\\frac{1-6b}{2b},\\frac{1-12b}{2b},….\\)
\nQuestion 3. <\/strong>How many two digit numbers are divisible by 3?
\nQuestion 4. <\/strong>What is the common difference of an A.P. in Which \\(a_{21}-a_7=84\\)?
\nQuestion 5.\u00a0<\/strong>If the sum of the first p<\/em> terms of an A.P. is \\(ap^2+bp,\\). Find its common difference.<\/p>\n
Two Marks Questions<\/strong><\/span><\/h3>\n
Question 1.<\/strong> If the \\(17^{th}\\) term of an A.P. exceeds its \\(10^{th}\\) term by 7, find the common difference.
\nQuestion 2.<\/strong> Find how many integers between 200 and 500 are divisible by 8.
\nQuestion 3.<\/strong> Which term of the progression \\(\\text{20,}19\\frac{1}{4},18\\frac{1}{2},17\\frac{3}{4},….\\) is the first negative term?
\nQuestion 4.<\/strong> Find the middle term of the AP \\(\\text{213,205,197,……,}37\\).
\nQuestion 5.<\/strong> Find whether -150 is a term of the A.P. \\(\\text{17,12,7,2,}…\\)?
\nQuestion 6.<\/strong> If the sum of the first p terms of an A.P. is \\(ap^2+bp,\\). Find its common difference.
\nQuestion 7.<\/strong> If \\(S_n\\), the sum of first n terms of an A.P. is given by \\(S_n=3n^2=4n,\\)find the \\(n^{\\text{th}}\\) term.
\nQuestion 8.<\/strong> In an A.P., if \\(S_5+S_7\\,\\,=\\,\\,\\text{167 and }S_{10}\\,\\,=\\,\\,\\text{235,}\\) then find the A.P., where denotes the sum of its first n terms.
\nQuestion 9.<\/strong>The sum of the first n terms of an A.P. is\\(5n\\,\\,-\\,\\,n^2.\\)Find the \\(n^{th}\\)term of this A.P.
\nQuestion 10.<\/strong> Find the common difference of an A.P. whose first term in 4, the last term is 49 and the sum of all its terms is 265.<\/p>\n
Three marks questions<\/strong><\/span><\/h3>\n
Question 1. <\/strong>If the \\(m^{th}\\)term of an A.P. is \\(\\frac{1}{n}\\) and \\(\\text{n}^{\\text{th}}\\) term is \\(\\frac{1}{m}\\) then show that its\\(\\left( mn \\right) ^{th}\\) term is 1.
\nQuestion 2.<\/strong> If the ratio of the sum of first n terms of two A.P’s is \\((7n + 1) : (4n + 27)\\), find the ratio of
\ntheir \\(m^{th}\\) terms.
\nQuestion 3.<\/strong> The digits of a positive number of three digits are in A.P. and their sum is 15. The number
\nobtained by reversing the digits is 594 less than the original number. Find the number.
\nQuestion 4.<\/strong> The\\(p^{\\text{th}}, q^{\\text{th}}\\,\\,\\text{and }r^{\\text{th}}\\) terms of an A.P. are a, b and c respectively. Show that \\(a\\left( q-\\text{r} \\right) +\\text{b}\\left( r-p \\right) +c\\left( p-q \\right) =0\\)
\nQuestion 5.<\/strong> Divide 56 in four parts in A.P. such that the ratio of the product of their extremes
\n\\(\\left( 1^{st}\\,\\,\\text{and }4^{th} \\right) \\)to the product of means \\(\\left( 2^{\\text{nd}}\\,\\,\\text{and }3^{\\text{rd}} \\right) \\) is 5:6.<\/p>\n
Four or six marks questions<\/strong><\/span><\/h3>\n
Question 1<\/strong> The sum of four consecutive numbers in an A.P. is 32 and the ratio of the product of the first and the last term to the product of two middle terms is \\(7: 15\\). Find the numbers.
\nQuestion 2.<\/strong> The \\(17^{\\text{th}}\\) term of an A.P. is 5 more than twice its \\(8^{\\text{th}}\\) term. If the \\(11^{th}\\) term of the A.P. is 43, then find its \\(n^{th}\\) term.<\/p>\n
Answers to one marks questions<\/strong>
\nAnswer 1<\/strong>
\nIt is given in the question that k, 2k – 1 and 2k + 1 are three consecutive terms of an A.P.
\nTherefore
\n\\(\\left( 2k-1 \\right) -\\left( k \\right) =\\left( 2k+1 \\right) -\\left( 2k-1 \\right) \\\\\\Rightarrow \\,\\,k-1=2\\\\\\Rightarrow k=3\\)
\nAnswer 2<\/strong>
\nThe common difference of the A.P. \\(\\frac{1}{2b},\\frac{1-6b}{2b},\\frac{1-12b}{2b},….\\) is given by
\n\\(\\frac{1-6b}{2b}-\\frac{1}{2b}=\\frac{1-6b-1}{2b}=\\frac{-6b}{2b}=-3\\)
\nAnswer 3<\/strong>
\nTwo-digit numbers which are divisible by 3 are 12,15,18 ,…, 99.
\nThis sequence forms an A.P. with first term \\(\\left( a \\right) =12\\),
\ncommon difference \\(\\left( d \\right) =15-12=3\\)
\nand last term \\(t_n=99\\)
\nWe know that
\n\\(t_n=a+\\left( n-1 \\right) d\\\\\\therefore \\,\\,12+\\left( n-1 \\right) 3=99\\\\\\Rightarrow 3n=99-9\\\\or, n=\\frac{90}{3}=30\\)
\nSo,\u00a0 there are\u00a0 30 two-digit numbers which\u00a0are divisible by 3.
\nAnswer 4<\/strong>
\nLet \\(a\\)\u00a0be the first term and \\(d\\)\u00a0be the common difference of the A.P.
\nGiven that,
\n\\(a_{21}-a_7=84\\\\t_n=a+\\left( n-1 \\right) d\\\\\\therefore \\,\\,a_{21}=a+20d\\,\\,and\\,\\,a_7=a+6d\\\\a+20d-\\left( a+6d \\right) =84\\\\\\Rightarrow 14d=84\\\\\\Rightarrow d=6\\)<\/p>\n
Answers to two marks questions<\/strong><\/p>\n
Answers to three marks questions<\/strong><\/p>\n
Answer 1<\/strong> Let \\(a\\) be the first term and \\(d\\) be the common difference of the given A.P.
\nNow, \\(r^{th}\\) term of A.P. \\(a_r=a+(r-1)d\\)
\nAccording to question,
\n\\(a_m =a+(m-1)d=\\frac{1}{n} \\qquad \\left( i \\right)\\)
\nand,
\n\\(a_n =a+\\left( n-1 \\right) d=\\frac{1}{m} \\qquad \\left( ii \\right)\\)
\nSubtracting, \\(\\left( ii \\right) \\text{from } \\left( i \\right)\\) , we get
\n\\((m – n)d = \\frac{m – n}{mn} \\Rightarrow d = \\frac{1}{mn}\\)
\nputting \\(d = \\frac{1}{mn}\\) in \\(\\left( i \\right)\\) , we get
\n\\(a + (m – n) \\frac{1}{mn} = \\frac{1}{n} \\Rightarrow a +\\frac{1}{n} – \\frac{1}{mn} = \\frac{1}{n}\\)
\n\\(\\Rightarrow a = \\frac{1}{mn}\\)
\n\\(\\therefore a_{mn} = a + (mn – 1)d = \\frac{1}{mn} + (mn – 1) \\frac{1}{mn}\\)
\n\\(= \\frac{1 + mn – 1}{mn} = 1\\)<\/p>\n
Answer 2<\/strong><\/p>\n
Let \\(a_1\\), \\(d_1\\), and \\(a_2\\), \\(d_2\\) be the first term and common difference of the two A.P.’s respectively.
\n\\(\\frac{n}{2} \\frac{ \\left[ 2a_1 + (n – 1)d_1 \\right] }{\\left[ 2a_2 + (n – 1)d_2 \\right]} = \\frac{7n + 1}{4n + 27}\u00a0 \\) as given in the question
\n\\(\\Rightarrow \\frac{a_1 + \\frac{\\left( n – 1 \\right)}{2}d_1}{a_2 + \\frac{\\left( n – 1 \\right)}{2}d_2} = \\frac{7n + 1}{4n + 27}\\)
\nput
\n\\(\\frac{n – 1}{2} = m – 1 \\Rightarrow n – 1 = 2m – 2\\)
\n\\(\\Rightarrow n = 2m – 2 + 1 = 2m – 1\\)
\n\\(\\therefore \\frac{a_1 + \\left( m-1 \\right) d_1}{a_2 + \\left( m-1 \\right) d_2}=\\frac{7\\left( 2m-1 \\right) +1}{4\\left( 2m-1 \\right) +27} \\)<\/p>\n
solving it further we get the ratio as,
\n\\( = \\frac{14m – 6}{8m – 4 + 27}\\)<\/p>\n
You can check also ncert solutions for class 10 maths<\/a> for practice<\/p>\n","protected":false},"excerpt":{"rendered":"
This page contains arithmetic progression questions for class 10. In this link get short arithmetic progression class 10 notes. Few other links are arithmetic progression class 10 important questions arithmetic progression assignment Practicing these questions on Arithmetic progression for class 10 maths can be helpful for your board exams. Arithmetic Progression questions and answers class […]<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[498],"tags":[],"class_list":["post-5257","post","type-post","status-publish","format-standard","hentry","category-maths"],"yoast_head":"\nArithmetic progression questions for class 10 - physicscatalyst's Blog<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/physicscatalyst.com\/article\/arithmetic-progression-questions-class-10\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Arithmetic progression questions for class 10 - physicscatalyst's Blog\" \/>\n<meta property=\"og:description\" content=\"This page contains arithmetic progression questions for class 10. 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