CBSE Class 11 Mathematics Chapter 9 Sequences and Series Multiple Choice Questions with Answers. MCQ Class 11 Mathematics Sequences and Series with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 11 Mathematics Sequences and Series MCQs with Answers to know their preparation level.
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MCQ Class 11 Mathematics Sequences and Series with Answers - Set - 1
Question 1:
If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/c, b/a, c/b are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) A.G.P.
Correct Answer – (C)
Given, equation is
ax² + bx + c = 0
Let p and q are the roots of this equation.
Now p+q = -b/a
and pq = c/a
Given that
p + q = 1/p² + 1/q²
⇒ p + q = (p² + q²)/(p² ×q²)
⇒ p + q = {(p + q)² – 2pq}/(pq)²
⇒ -b/a = {(-b/a)² – 2c/a}/(c/a)²
⇒ (-b/a)×(c/a)² = {b²/a² – 2c/a}
⇒ -bc²/a³ = {b² – 2ca}/a²
⇒ -bc²/a = b² – 2ca
Divide by bc on both side, we get
⇒ -c /a = b/c – 2a/b
⇒ 2a/b = b/c + c/a
⇒ b/c, a/b, c/a are in AP
⇒ c/a, a/b, b/c are in AP
⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP
⇒ a/c, b/a, c/b are in HP
Question 2 :
Let Tr be the r th term of an A.P., for r = 1, 2, 3, … If for some positive integers m, n, we have Tm = 1/n and Tn = 1/m, then Tm n equals
(a) 1/m n
(b) 1/m + 1/n
(c) 1
(d) 0
Correct Answer – (C)
Let first term is a and the common difference is d of the AP
Now, Tm = 1/n
⇒ a + (m-1)d = 1/n ………… 1
and Tn = 1/m
⇒ a + (n-1)d = 1/m ………. 2
From equation 2 – 1, we get
(m-1)d – (n-1)d = 1/n – 1/m
⇒ (m-n)d = (m-n)/mn
⇒ d = 1/mn
From equation 1, we get
a + (m-1)/mn = 1/n
⇒ a = 1/n – (m-1)/mn
⇒ a = {m – (m-1)}/mn
⇒ a = {m – m + 1)}/mn
⇒ a = 1/mn
Now, Tmn = 1/mn + (mn-1)/mn
⇒ Tmn = 1/mn + 1 – 1/mn
⇒ Tmn = 1
Question 3 :
The sum of series 1/2! + 1/4! + 1/6! + ….. is
(a) e² – 1 / 2
(b) (e – 1)² /2 e
(c) e² – 1 / 2 e
(d) e² – 2 / e
Correct Answer – (B)
We know that,
ex = 1 + x/1! + x² /2! + x³ /3! + x4 /4! + ………..
Now,
e1 = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ………..
e-1 = 1 – 1/1! + 1/2! – 1/3! + 1/4! + ………..
e1 + e-1 = 2(1 + 1/2! + 1/4! + ………..)
⇒ e + 1/e = 2(1 + 1/2! + 1/4! + ………..)
⇒ (e² + 1)/e = 2(1 + 1/2! + 1/4! + ………..)
⇒ (e² + 1)/2e = 1 + 1/2! + 1/4! + ………..
⇒ (e² + 1)/2e – 1 = 1/2! + 1/4! + ………..
⇒ (e² + 1 – 2e)/2e = 1/2! + 1/4! + ………..
⇒ (e – 1)² /2e = 1/2! + 1/4! + ………..
Question 4 :
The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is
(a) n/(n+1)
(b) 1/(n+1)
(c) 1/n
(d) None of these
Correct Answer – (A)
Given series is:
S = (1/1·2) + (1/2·3) + (1/3·4) – ………………. 1/n.(n+1)
⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -……… (1/n – 1/(n+1))
⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)
⇒ S = 1 – 1/(n+1)
⇒ S = (n + 1 – 1)/(n+1)
⇒ S = n/(n+1)
Question 5 :
If a, b, c are in AP then
(a) b = a + c
(b) 2b = a + c
(c) b² = a + c
(d) 2b² = a + c
Correct Answer – (B)
Given, a, b, c are in AP
⇒ b – a = c – b
⇒ b + b = a + c
⇒ 2b = a + c
MCQ Class 11 Mathematics Sequences and Series with Answers
Question 6 :
The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is
(a) 2
(b) 4
(c) 6
(d) 8
Correct Answer – (C)
Let a and b are two numbers such that
a + b = 13/6
Let A1, A2, A3, ………A2n be 2n arithmetic means between a and b
Then, A1 + A2 + A3 + ………+ A2n = 2n{(n + 1)/2}
⇒ n(a + b) = 13n/6
Given that A1 + A2 + A3 + ………+ A2n = 2n + 1
⇒ 13n/6 = 2n + 1
⇒ n = 6
Question 7 :
The third term of a geometric progression is 4. The product of the first five terms is
(a) 43
(b) 45
(c) 44
(d) none of these
Correct Answer – (B)
here it is given that T3 = 4.
⇒ ar² = 4
Now product of first five terms = a.ar.ar².ar³.ar4
= a5r10
= (ar2)5
= 45
Question 8 :
If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then
(a) a, b, c are in AP
(b) a², b², c² are in AP
(c) 1/1, 1/b, 1/c are in AP
(d) None of these
Correct Answer – (B)
Given, 1/(b + c), 1/(c + a), 1/(a + b)
⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)
⇒ 2b² = a² + c²
⇒ a², b², c² are in AP
Question 9 :
Three numbers form an increasing GP. If the middle term is doubled, then the new numbers are in Ap. The common ratio of GP is
(a) 2 + √3
(b) 2 – √3
(c) 2 ± √3
(d) None of these
Correct Answer – (A)
Let the three numbers be a/r, a, ar
Since the numbers form an increasing GP, So r > 1
Now, it is given that a/r, 2a, ar are in AP
⇒ 4a = a/r + ar
⇒ r² – 4r + 1 = 0
⇒ r = 2 ± √3
⇒ r = 2 + √3 {Since r > 1}
Question 10 :
If a, b, c are in G.P., then the equations ax² + 2bx + c = 0 and dx² + 2ex + f = 0 have a common root if d/a, e/b, f/c are in
(a) AP
(b) GP
(c) HP
(d) none of these
Correct Answer – (A)
Given a, b, c are in GP
⇒ b² = ac
⇒ b² – ac = 0
So, ax² + 2bx + c = 0 have equal roots.
Now D = 4b² – 4ac
and the root is -2b/2a = -b/a
So -b/a is the common root.
Now,
dx² + 2ex + f = 0
⇒ d(-b/a)² + 2e×(-b/a) + f = 0
⇒ db2 /a² – 2be/a + f = 0
⇒ d×ac /a² – 2be/a + f = 0
⇒ dc/a – 2be/a + f = 0
⇒ d/a – 2be/ac + f/c = 0
⇒ d/a + f/c = 2be/ac
⇒ d/a + f/c = 2be/b²
⇒ d/a + f/c = 2e/b
⇒ d/a, e/b, f/c are in AP
- NCERT Solutions Class 11 Mathematics Sequences and Series with Answers : Exercise 9.1
- NCERT Solutions Class 11 Mathematics Sequences and Series with Answers : Exercise 9.2
- NCERT Solutions Class 11 Mathematics Sequences and Series with Answers : Exercise 9.3
- NCERT Solutions Class 11 Mathematics Sequences and Series with Answers : Exercise 9.4
- NCERT Solutions Class 11 Mathematics Sequences and Series with Answers : Exercise 9 Misc
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