{"id":17479,"date":"2022-05-24T22:00:59","date_gmt":"2022-05-24T16:30:59","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17479"},"modified":"2022-05-16T15:43:38","modified_gmt":"2022-05-16T10:13:38","slug":"mcq-questions-for-class-11-maths-chapter-9","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-maths-chapter-9\/","title":{"rendered":"MCQ Questions for Class 11 Maths Chapter 9 \ufeffSequences and Series with Answers"},"content":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 9 Sequences and Series with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers<\/a> during preparation and score maximum marks in the exam. Students can download the Sequences and Series Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 9 Sequences and Series Objective Questions.<\/p>\n

Sequences and Series Class 11 MCQs Questions with Answers<\/h2>\n

Students are advised to solve the Sequences and Series Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Sequences and Series Class 11 with answers will boost your confidence thereby helping you score well in the exam.<\/p>\n

Explore numerous MCQ Questions of Sequences and Series Class 11 with answers provided with detailed solutions by looking below.<\/p>\n

Question 1.
\nLet Tr be the rth term of an A.P. whose first term is a and the common difference is d. If for some positive integers m, n, m \u2260 n, T m = 1\/n and T n = 1\/m then (a-d) equals to
\n(a) 0
\n(b) 1
\n(c) 1\/mn
\n(d) 1\/m + 1\/n<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 0
\nGiven the first term is a and the common difference is d of the AP
\nNow, Tm = 1\/n
\n\u21d2 a + (m – 1)d = 1\/n ………… 1
\nand Tn = 1\/m
\n\u21d2 a + (n – 1)d = 1\/m ………. 2
\nFrom equation 2 – 1, we get
\n(m – 1)d – (n – 1)d = 1\/n – 1\/m
\n\u21d2 (m – n)d = (m – n)\/mn
\n\u21d2 d = 1\/mn
\nFrom equation 1, we get
\na + (m – 1)\/mn = 1\/n
\n\u21d2 a = 1\/n – (m – 1)\/mn
\n\u21d2 a = {m – (m – 1)}\/mn
\n\u21d2 a = {m – m + 1)}\/mn
\n\u21d2 a = 1\/mn
\nNow, a – d = 1\/mn – 1\/mn
\n\u21d2 a – d = 0<\/p>\n<\/details>\n

\n

Question 2.
\nThe first term of a GP is 1. The sum of the third term and fifth term is 90. The common ratio of GP is
\n(a) 1
\n(b) 2
\n(c) 3
\n(d) 4<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 3
\nLet first term of the GP is a and common ratio is r.
\n3rd term = ar\u00b2
\n5th term = ar4<\/sup>
\nNow
\n\u21d2 ar\u00b2 + ar4<\/sup> = 90
\n\u21d2 a(r\u00b2 + r4<\/sup>) = 90
\n\u21d2 r\u00b2 + r4<\/sup> = 90
\n\u21d2 r\u00b2 \u00d7 (r\u00b2 + 1) = 90
\n\u21d2 r\u00b2 (r\u00b2 + 1) = 3\u00b2 \u00d7 (3\u00b2 + +1)
\n\u21d2 r = 3
\nSo the common ratio is 3<\/p>\n<\/details>\n

\n

Question 3.
\nIf a is the first term and r is the common ratio then the nth term of GP is
\n(a) (ar)n-1<\/sup>
\n(b) a \u00d7 r\u207f
\n(c) a \u00d7 rn-1<\/sup>
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) a \u00d7 rn-1<\/sup>
\nGiven, a is the first term and r is the common ratio.
\nNow, nth term of GP = a \u00d7 rn-1<\/sup><\/p>\n<\/details>\n

\n

Question 4.
\nThe sum of odd integers from 1 to 2001 is
\n(a) 10201
\n(b) 102001
\n(c) 100201
\n(d) 1002001<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 1002001
\nThe odd numbers from 1 to 2001 are:
\n1, 3, 5, ………, 2001
\nThis froms an AP
\nwhere first term a = 1
\nCommon difference d = 3 – 1 = 2
\nlast term l = 2001
\nLet number of terms = n
\nNow, l = a + (n – 1)d
\n\u21d2 2001 = 1 + (n – 1)2
\n\u21d2 2001 – 1 = (n – 1)2
\n\u21d2 2(n – 1) = 2000
\n\u21d2 n – 1 = 2000\/2
\n\u21d2 n – 1 = 1000
\n\u21d2 n = 1001
\nNow, sum = (n\/2) \u00d7 (a + 1)
\n= (1001\/2) \u00d7 (1 + 2001)
\n= (1001\/2) \u00d7 2002
\n= 1001 \u00d7 1001
\n= 1002001
\nSo, the sum of odd integers from 1 to 2001 is 1002001<\/p>\n<\/details>\n

\n

Question 5.
\nIf a, b, c are in AP and x, y, z are in GP then the value of xb-c<\/sup> \u00d7 yc-a<\/sup> \u00d7 za-b<\/sup> is
\n(a) 0
\n(b) 1
\n(c) -1
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 1
\nGiven, a, b, c are in AP
\n\u21d2 2b = a + c ………. 1
\nand x, y, z are in GP
\n\u21d2 y\u00b2 = xz ……….. 2
\nNow, xb-c<\/sup> \u00d7 yc-a<\/sup> \u00d7 za-b<\/sup> = xb-c<\/sup> \u00d7 (\u221axz)c-a<\/sup> \u00d7 za-b<\/sup>
\n= xb-c<\/sup> \u00d7 x(c-a)\/2<\/sup> \u00d7 z(c-a)\/2<\/sup> \u00d7 za-b<\/sup>
\n= xb-c<\/sup> + x(c-a)\/2<\/sup> \u00d7 z(c-a)\/2+ a -b<\/sup>
\n= x2b+(c+a)<\/sup> \u00d7 z(c+a)-2b<\/sup>
\n= x\u00b0 \u00d7 z\u00b0
\n= 1
\nSo, the value of xb-c<\/sup> \u00d7 yc-a<\/sup> \u00d7 za-b<\/sup> is 1<\/p>\n<\/details>\n

\n

Question 6.
\nAn example of geometric series is
\n(a) 9, 20, 21, 28
\n(b) 1, 2, 4, 8
\n(c) 1, 2, 3, 4
\n(d) 3, 5, 7, 9<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 1, 2, 4, 8
\n1, 2, 4, 8 is the example of geometric series
\nHere common ratio = 2\/1 = 4\/2 = 8\/4 = 2<\/p>\n<\/details>\n

\n

Question 7.
\nThree numbers from an increasing GP of the middle number is doubled, then the new numbers are in AP. The common ratio of the GP is
\n(a) 2
\n(b) \u221a3
\n(c) 2 + \u221a3
\n(d) 2 – \u221a3<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 2 + \u221a3
\nGiven that three numbers from an increasing GP
\nLet the 3 number are: a, ar, ar\u00b2 (r > 1)
\nNow, according to question,
\na, 2ar, ar\u00b2 are in AP
\nSo, 2ar – a = ar\u00b2 – 2ar
\n\u21d2 a(2r – 1) = a(r\u00b2 – 2r)
\n\u21d2 2r – 1 = r\u00b2 – 2r
\n\u21d2 r\u00b2 – 2r – 2r + 1 = 0
\n\u21d2 r\u00b2 – 4r + 1 = 0
\n\u21d2 r = [4 \u00b1 \u221a{16 – 4 \u00d7 1 \u00d7 1}]\/2
\n\u21d2 r = [4 \u00b1 \u221a{16 – 4}]\/2
\n\u21d2 r = {4 \u00b1 \u221a12}\/2
\n\u21d2 r = {4 \u00b1 2\u221a3}\/2
\n\u21d2 r = {2 \u00b1 \u221a3}
\nSince r > 1
\nSo, the common ratio of the GP is (2 + \u221a3)<\/p>\n<\/details>\n

\n

Question 8.
\nAn arithmetic sequence has its 5th term equal to 22 and its 15th term equal to 62. Then its 100th term is equal to
\n(a) 410
\n(b) 408
\n(c) 402
\n(d) 404<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 402
\nLet ais the first term and d is the common difference of the AP
\nGiven,
\na5<\/sub> = a + (5 – 1)d = 22
\n\u21d2 a + 4d = 22 ………….1
\nand a15 = a + (15 – 1)d = 62
\n\u21d2 a + 14d = 62 ………2
\nFrom equation 2 – 1, we get
\n62 – 22 = 14d – 4d
\n\u21d2 10d = 40
\n\u21d2 d = 4
\nFrom equation 1, we get
\na + 4 \u00d7 4 = 22
\n\u21d2 a + 16 = 22
\n\u21d2 a = 6
\nNow,
\na100 = 6 + 4(100 – 1 )
\n\u21d2 a100 = 6 + 4 \u00d7 99
\n\u21d2 a100 = 6 + 396
\n\u21d2 a100 = 402<\/p>\n<\/details>\n

\n

Question 9.
\nSuppose a, b, c are in A.P. and a\u00b2, b\u00b2, c\u00b2 are in G.P. If a < b < c and a + b + c = 3\/2, then the value of a is
\n(a) 1\/2\u221a2
\n(b) 1\/2\u221a3
\n(c) 1\/2 – 1\/\u221a3
\n(d) 1\/2 – 1\/\u221a2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 1\/2 – 1\/\u221a2
\nGiven, a, b, c are in AP
\n\u21d2 2b = a + c
\n\u21d2 b = (a + c)\/2 ………….. 1
\nAgain given, a\u00b2, b\u00b2, c\u00b2 are in GP then b4<\/sup> = a\u00b2 c\u00b2
\n\u21d2 b\u00b2 = \u00b1 ac \u2026……… 2
\nUsing 1 in a + b + c = 3\/2, we get
\n3b = 3\/2
\n\u21d2 b = 1\/2
\nhence a + c = 1
\nand ac = \u00b1 1\/4
\nSo a & c are roots of either x2 \u2212x + 1\/4 = 0 or x\u00b2 \u2212 x \u2212 1\/4 = 0
\nThe first has equal roots of x = 1\/2 and second gives x= (1 \u00b1 \u221a2)\/2 for a and c
\nSince a < c,
\nwe must have a = (1\u2212\u221a2)\/2
\n\u21d2 a – 1\/2 – \u221a2\/2
\n\u21d2 a – 1\/2 – \u221a2\/(\u221a2\u00d7\u221a2)
\n\u21d2 a – 1\/2 – 1\/\u221a2<\/p>\n<\/details>\n

\n

Question 10.
\nIf the positive numbers a, b, c, d are in A.P., then abc, abd, acd, bcd are
\n(a) not in A.P. \/ G.P. \/ H. P.
\n(b) in A.P.
\n(c) in G.P.
\n(d) in H.P.<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) in H.P.
\nGiven, the positive numbers a, b, c, d are in A.P.
\n\u21d2 1\/a, 1\/b, 1\/c, 1\/d are in H.P.
\n\u21d2 1\/d, 1\/c, 1\/b, 1\/a are in H.P.
\nNow, Multiply by abcd, we get
\nabcd\/d, abcd\/c, abcd\/b, abcd\/a are in H.P.
\n\u21d2 abc, abd, acd, bcd are in H.P.<\/p>\n<\/details>\n

\n

Question 11.
\nLet Tr be the rth term of an A.P. whose first term is a and the common difference is d. If for some positive integers m, n, m \u2260 n, T m = 1\/n and T n = 1\/m then (a-d) equals to
\n(a) 0
\n(b) 1
\n(c) 1\/mn
\n(d) 1\/m + 1\/n<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 0
\nGiven the first term is a and the common difference is d of the AP
\nNow, Tm = 1\/n
\n\u21d2 a + (m – 1)d = 1\/n ………… 1
\nand Tn = 1\/m
\n\u21d2 a + (n – 1)d = 1\/m ………. 2
\nFrom equation 2 – 1, we get
\n(m – 1)d – (n – 1)d = 1\/n – 1\/m
\n\u21d2 (m – n)d = (m – n)\/mn
\n\u21d2 d = 1\/mn
\nFrom equation 1, we get
\na + (m – 1)\/mn = 1\/n
\n\u21d2 a = 1\/n – (m – 1)\/mn
\n\u21d2 a = {m – (m – 1)}\/mn
\n\u21d2 a = {m – m + 1)}\/mn
\n\u21d2 a = 1\/mn
\nNow, a – d = 1\/mn – 1\/mn
\n\u21d2 a – d = 0<\/p>\n<\/details>\n

\n

Question 12.
\nIn the sequence obtained by omitting the perfect squares from the sequence of natural numbers, then 2011th term is
\n(a) 2024
\n(b) 2036
\n(c) 2048
\n(d) 2055<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 2055
\nBefore 2024, there are 44 squares,
\nSo, 1980th term is 2024
\nHence, 2011th term is 2055<\/p>\n<\/details>\n

\n

Question 13.
\nIf the first term minus third term of a G.P. = 768 and the third term minus seventh term of the same G.P. = 240, then the product of first 21 terms =
\n(a) 1
\n(b) 2
\n(c) 3
\n(d) 4<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 1
\nLet first term = a
\nand common ratio = r
\nGiven, a – ar\u00b2 = 768
\n\u21d2 a(1 – r\u00b2) = 768
\nand ar\u00b2 – ar6<\/sup> = 240
\n\u21d2 ar\u00b2 (1 – r4<\/sup>) = 240
\nDividing the above 2 equations, we get
\nar\u00b2 (1 – r4<\/sup>)\/a(1 – r\u00b2) = 240\/768
\n\u21d2 {ar\u00b2 (1 – r\u00b2) \u00d7 (1 + r\u00b2)}\/a(1 – r\u00b2) = 240\/768
\n\u21d2 1 + r\u00b2 = 0.3125
\n\u21d2 r\u00b2 = 0.25
\n\u21d2 r\u00b2 = 25\/100
\n\u21d2 r\u00b2 = \u221a(1\/4)
\n\u21d2 r = \u00b1 1\/2
\nNow, a(1 – r\u00b2) = 768
\n\u21d2 a(1 – 1\/4 ) = 768
\n\u21d2 3a\/4 = 768
\n\u21d2 3a = 4 \u00d7 768
\n\u21d2 a = (4 \u00d7 768)\/3
\n\u21d2 a = 4 \u00d7 256
\n\u21d2 a = 1024
\n\u21d2 a = 210<\/sup>
\nNow product of first 21 terms = (a\u00b2 \u00d7 r20<\/sup>)10<\/sup> \u00d7 a \u00d7 r10<\/sup>
\n= a21<\/sup> \u00d7 r210<\/sup>
\n= (210<\/sup>)21<\/sup> \u00d7 (1\/2)210<\/sup>
\n= 2210<\/sup> \/2210<\/sup>
\n= 1<\/p>\n<\/details>\n

\n

Question 14.
\nIf the sum of the first 2n terms of the A.P. 2, 5, 8, ….., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ….., then n equals
\n(a) 10
\n(b) 12
\n(c) 11
\n(d) 13<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 11
\nGiven, the sum of the first 2n terms of the A.P. 2, 5, 8, ….. = the sum of the first n terms of the A.P. 57, 59, 61, ….
\n\u21d2 (2n\/2) \u00d7 {2 \u00d7 2 + (2n – 1)3} = (n\/2) \u00d7 {2 \u00d7 57 + (n – 1)2}
\n\u21d2 n \u00d7 {4 + 6n – 3} = (n\/2) \u00d7 {114 + 2n – 2}
\n\u21d2 6n + 1 = {2n + 112}\/2
\n\u21d2 6n + 1 = n + 56
\n\u21d2 6n – n = 56 – 1
\n\u21d2 5n = 55
\n\u21d2 n = 55\/5
\n\u21d2 n = 11<\/p>\n<\/details>\n

\n

Question 15.
\nIf a, b, c are in GP then log a\u207f, log b\u207f, log c\u207f are in
\n(a) AP
\n(b) GP
\n(c) Either in AP or in GP
\n(d) Neither in AP nor in GP<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) AP
\nGiven, a, b, c are in GP
\n\u21d2 b\u00b2 = ac
\n\u21d2 (b\u00b2)\u207f = (ac)\u207f
\n\u21d2 (b2 )\u207f= a\u207f \u00d7 c\u207f
\n\u21d2 log (b\u00b2)\u207f = log(an \u00d7 cn )
\n\u21d2 log b\u00b2\u207f = log a\u207f + log c\u207f
\n\u21d2 log (b\u207f)\u00b2 = log a\u207f + log c\u207f
\n\u21d2 2 \u00d7 log b\u207f = log a\u207f + log c\u207f
\n\u21d2 log a\u207f, log b\u207f, log c\u207f are in AP<\/p>\n<\/details>\n

\n

Question 16.
\nIf the nth term of an AP is 3n – 4, the 10th term of AP is
\n(a) 12
\n(b) 22
\n(c) 28
\n(d) 30<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 28
\nGiven, an<\/sub> = 3n – 2
\nPut n = 10, we get
\na10<\/sub> = 3 \u00d7 10 – 2
\n\u21d2 a10<\/sub> = 30 – 2
\n\u21d2 a10<\/sub> = 28
\nSo, the 10th term of AP is 28<\/p>\n<\/details>\n

\n

Question 17.
\nIf the third term of an A.P. is 7 and its 7 th term is 2 more than three times of its third term, then the sum of its first 20 terms is
\n(a) 228
\n(b) 74
\n(c) 740
\n(d) 1090<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 740
\nLet a is the first term and d is the common difference of AP
\nGiven the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term
\n\u21d2 a + 2d = 7 ………….. 1
\nand
\n3(a + 2d) + 2 = a + 6d
\n\u21d2 3 \u00d7 7 + 2 = a + 6d
\n\u21d2 21 + 2 = a + 6d
\n\u21d2 a + 6d = 23 ………….. 2
\nFrom equation 1 – 2, we get
\n4d = 16
\n\u21d2 d = 16\/4
\n\u21d2 d = 4
\nFrom equation 1, we get
\na + 2 \u00d7 4 = 7
\n\u21d2 a + 8 = 7
\n\u21d2 a = -1
\nNow, the sum of its first 20 terms
\n= (20\/2) \u00d7 {2 \u00d7 (-1) + (20-1) \u00d7 4}
\n= 10 \u00d7 {-2 + 19 \u00d7 4)}
\n= 10 \u00d7 {-2 + 76)}
\n= 10 \u00d7 74
\n= 740<\/p>\n<\/details>\n

\n

Question 18.
\nIf a, b, c are in AP then
\n(a) b = a + c
\n(b) 2b = a + c
\n(c) b\u00b2 = a + c
\n(d) 2b\u00b2 = a + c<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 2b = a + c
\nGiven, a, b, c are in AP
\n\u21d2 b – a = c – b
\n\u21d2 b + b = a + c
\n\u21d2 2b = a + c<\/p>\n<\/details>\n

\n

Question 19.
\nIf 1\/(b + c), 1\/(c + a), 1\/(a + b) are in AP then
\n(a) a, b, c are in AP
\n(b) a\u00b2, b\u00b2, c\u00b2 are in AP
\n(c) 1\/1, 1\/b, 1\/c are in AP
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) a\u00b2, b\u00b2, c\u00b2 are in AP
\nGiven, 1\/(b + c), 1\/(c + a), 1\/(a + b)
\n\u21d2 2\/(c + a) = 1\/(b + c) + 1\/(a + b)
\n\u21d2 2b2 = a\u00b2 + c\u00b2
\n\u21d2 a\u00b2, b\u00b2, c\u00b2 are in AP<\/p>\n<\/details>\n

\n

Question 20.
\n3, 5, 7, 9, …….. is an example of
\n(a) Geometric Series
\n(b) Arithmetic Series
\n(c) Rational Exponent
\n(d) Logarithm<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) Arithmetic Series
\n3, 5, 7, 9, …….. is an example of Arithmetic Series.
\nHere common difference = 5 – 3 = 7 – 5 = 9 – 7 = 2<\/p>\n<\/details>\n

\n

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 9 Sequences and Series with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can …<\/p>\n

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