{"id":2980,"date":"2023-11-28T04:12:24","date_gmt":"2023-11-27T22:42:24","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=2980"},"modified":"2023-11-29T12:34:45","modified_gmt":"2023-11-29T07:04:45","slug":"samacheer-kalvi-11th-maths-solutions-chapter-5-ex-5-3","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-11th-maths-solutions-chapter-5-ex-5-3\/","title":{"rendered":"Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3"},"content":{"rendered":"

You can Download Samacheer Kalvi 11th Maths Book Solutions<\/a> Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3<\/h2>\n

Question 1.
\nFind the sum of the first 20-terms of the arithmetic progression having the sum of first 10 terms as 52 and the sum of the first 15 terms as 77.
\nSolution:
\nLet ‘a’ be the first term and d be the common difference of A.P.
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 2.
\n\"Samacheer
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 3.
\nCompute the sum of first n terms of the following series:
\n(i) 8 + 88 + 888 + 8888 + ……
\n(i) 6 + 66 + 666 + 6666 + …..
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 4.
\nCompute the sum of first n terms of 1 + (1 + 4) + (1 + 4 + 42<\/sup>) + (1 + 4 + 42<\/sup> + 43<\/sup>) + ……..
\nSolution:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nFind the general term and sum to n terms of the sequence \\(1, \\frac{4}{3}, \\frac{7}{9}, \\frac{10}{27}, \\ldots \\ldots \\ldots\\)
\nSolution:
\n\"Samacheer
\nDenominator 1, 3, 9, 27, which is a G.P. a = 1, r = 3
\n\"Samacheer
\nIt is an arithmetico Geometric series. Here the nth<\/sup> term is tn<\/sub> = [a + (n – 1)d]rn – 1<\/sup> where a = 1, d = 3 and r = 1\/3
\nNow the sum to n terms is
\n\"Samacheer<\/p>\n

Question 6.
\n\"Samacheer
\nSolution:
\n\"Samacheer<\/p>\n

Question 7.
\nShow that the sum of (m + n)th<\/sup> and (m – n)th<\/sup> term of an A.P. is equal to twice the mth<\/sup> term.
\nSolution:
\nLet the A.P. be a, a + d, a + 2d, ……..
\ntm + n<\/sub> = a + (m + n – 1)d
\ntm – n<\/sub> = a + (m – n – 1)d
\ntm<\/sub> = a + (m – 1)d
\n2tm<\/sub> = 2[a + (m – 1)d]
\nTo prove tm + n<\/sub> + tm – n<\/sub> = 2tm<\/sub>
\nLHS tm + n<\/sub> + tm – n<\/sub> = a + (m + n – 1)d + a + (m – n – 1)d
\n\"Samacheer
\n= 2a + d [2m – 2]
\n= 2[a + (m – 1)d) = 2 tm<\/sub> = RHS<\/p>\n

Question 8.
\nA man repays an amount of \u20b9 3250 by paying \u20b9 20 in the first month and then increases the payment by \u20b9 15 per month. How long will it take him to clear the amount?
\nSolution:
\na = 20, d = 15, Sn<\/sub> = 3250 to find n.
\n\"Samacheer
\n3n2<\/sup> + 5n – 1300 = 0
\n3n2<\/sup> – 60n + 65n – 1300 = 0
\n3n (n – 20) + 65 (n – 20) = 0
\n(3n + 65) (n – 20) = 0
\nn = – 65\/3 or 20
\nn = – 65\/3 is not possible
\n\u2234 n = 20
\nSo he will take 20 months to clear the amount.<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\nIn a race, 20 balls are placed in a line at intervals of 4 meters with the first ball 24 meters away from the starting point. A contestant is required to bring the balls back to the starting place one at a time. How far would the contestant run to bring back all balls?
\nSolution:
\nt1<\/sub> = 24 \u00d7 2 = 48, t2<\/sub> =48 + 8 = 56 or (24 + 4)2, t3<\/sub> =(28 + 4)2 = 64 which is an A.P.
\nHere a = 48,
\nd = 56 – 48 = 8
\n\"Samacheer
\nThe contestant has to run 2480 m to bring all the balls.<\/p>\n

Question 10.
\nThe number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd<\/sup> hour, 4th<\/sup> hour and nth<\/sup> hour?
\nSolution:
\nNumber of bacteria at the beginning = 30
\nNumber of bacteria after 1 hour = 30 \u00d7 2 = 60
\nNumber of bacteria after 2 hours = 30 \u00d7 22<\/sup> = 120
\nNumber of bacteria after 4 hours = 30 \u00d7 24 = 30 \u00d7 16 = 480
\n\u2234 Number of bacteria after nth<\/sup> hour = 30 \u00d7 2n<\/sup><\/p>\n

Question 11.
\nWhat will \u20b9 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
\nSolution:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 12.
\nIn a certain town, a viral disease caused severe health hazards upon its people disturbing their normal life. It was found that on each day, the virus which caused the disease spread in Geometric Progression. The amount of infectious virus particle gets doubled each day, being 5 particles on the first day. Find the day when the infectious virus particles just grow over 1,50,000 units?
\nSolution:
\na = 5, r = 2, tn<\/sub> >150000, To find ‘n’
\n\"Samacheer
\nOn the 15th<\/sup> day it will grow over 150000 units.<\/p>\n

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 Additional Questions<\/h3>\n

Question 1.
\nIf the ratio of the sums of m terms and n terms of an A.P. be m2<\/sup> : n2<\/sup>, prove that the ratio of its mth<\/sup> and nth<\/sup> terms is (2m – 1) : (2n – 1).
\nSolution:
\nLet ‘a’ be the first term and d be the common difference of A.P.
\n\"Samacheer
\n\u21d2 2an + (mn – n)d = 2am + (mn – m)d
\n\u21d2 2an – 2am = (mn – m – mn + n)d
\n\u21d2 2a(n – m) = (n – m)d \u21d2 d = 2a, [n – m \u2260 0, as n \u2260 m]
\n\"Samacheer
\nHence, the required ratio is (2m – 1) : (2n – 1)<\/p>\n

Question 2.
\nDetermine the number of terms of geometric progression {an<\/sub>} if a1<\/sub> = 3, an<\/sub> = 96, Sn<\/sub> = 189.
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 3.
\nThe sum of first three terms of a G.P. is to the sum of the first six terms as 125 : 152. Find the common ratio of the G.P.
\nSolution:
\n\"Samacheer
\nHence, the common ratio of the G.P. is \\(\\frac{3}{5}\\)<\/p>\n

Question 4.
\nFind the sum to n terms the series: (x + y) + (x2<\/sup> + xy + y2<\/sup>) + (x3<\/sup> + x2<\/sup>y + xy2<\/sup> + y3<\/sup>) + …
\nSolution:
\n(x + y) + (x2<\/sup> + xy + y2<\/sup>) + (x3<\/sup> + x2<\/sup>y + xy2<\/sup> + y3<\/sup>) + … to n terms
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 5.
\nSum the series: (1 + x) + (1 + x + x2<\/sup>) + (1 + x + x2<\/sup> + x3<\/sup>) + … up to n terms
\nSolution:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nSum up to n terms the series: 7 + 77 + 777 + 7777 + …
\nSolution:
\nS = 1 + 77 + 777 + 7777 + … to n terms
\n\"Samacheer<\/p>\n

Question 7.
\nIf S1<\/sub>, S2<\/sub>, S3<\/sub> be respectively the sums of n, 2n, 3n, terms of a G.P. , then prove that S1<\/sub>(S3<\/sub> – S2<\/sub>) = (S2<\/sub> – S1<\/sub>)2<\/sub>.
\nSolution:
\nLet a be the first term and r be the common ratio of G.P.
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 8.
\nIf sum of the n terms of a G.P be S, their product P and the sum of their reciprocals R, the prove that \\(P^{2}=\\left(\\frac{S}{R}\\right)^{n}\\)
\nSolution:
\nLet a be the first term and r be the common ratio of the G.P.
\n\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 Question 1. Find the sum of the first 20-terms of the arithmetic …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[7],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/2980"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=2980"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/2980\/revisions"}],"predecessor-version":[{"id":60023,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/2980\/revisions\/60023"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=2980"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=2980"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=2980"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}