CBSE Class 11 Maths – MCQ and Online Tests – Unit 9 – Sequences and Series
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CBSE Class 11 Maths – MCQ and Online Tests – Unit 9 – Sequences and Series
Question 1.
The sum of AP 2, 5, 8, …..up to 50 terms is
(a) 3557
(b) 3775
(c) 3757
(d) 3575
Answer
Answer: (b) 3775
Hint:
Given, AP is 2, 5, 8, …..up to 50
Now, first term a = 2
common difference d = 5 – 2 = 3
Number of terms = 50
Now, Sum = (n/2)Ũ{2a + (n – 1)d}
= (50/2)Ũ{2Ũ2 + (50 – 1)3}
= 25Ũ{4 + 49Ũ3}
= 25Ũ(4 + 147)
= 25 Ũ 151
= 3775
Question 2.
If 2/3, k, 5/8 are in AP then the value of k is
(a) 31/24
(b) 31/48
(c) 24/31
(d) 48/31
Answer
Answer: (b) 31/48
Hint:
Given, 2/3, k, 5/8 are in AP
? 2k = 2/3 + 5/8
? 2k = 31/24
? k = 31/48
So, the value of k is 31/48
Question 3:
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 + v3
(b) 2 – v3
(c) 2 ą v3
(d) None of these
Answer
Answer: (a) 2 + v3
Hint:
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
? rē – 4r + 1 = 0
? r = 2 ą v3
? r = 2 + v3 {Since r > 1}
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
Answer
Answer: (a) n/(n+1)
Hint:
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 1/(b + c), 1/(c + a), 1/(a + b) are in AP then
(a) a, b, c are in AP
(b) aē, bē, cē are in AP
(c) 1/1, 1/b, 1/c are in AP
(d) None of these
Answer
Answer: (b) aē, bē, cē are in AP
Hint:
Given, 1/(b + c), 1/(c + a), 1/(a + b)
? 2/(c + a) = 1/(b + c) + 1/(a + b)
? 2bē = aē + cē
? aē, bē, cē are in AP
Question 6:
The sum of series 1/2! + 1/4! + 1/6! + ….. is
(a) eē – 1 / 2
(b) (e – 1)ē /2 e
(c) eē – 1 / 2 e
(d) eē – 2 / e
Answer
Answer: (b) (e – 1)ē /2 e
Hint:
We know that,
ex = 1 + x/1! + xē /2! + xģ /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! + ………..)
? (eē + 1)/e = 2(1 + 1/2! + 1/4! + ………..)
? (eē + 1)/2e = 1 + 1/2! + 1/4! + ………..
? (eē + 1)/2e – 1 = 1/2! + 1/4! + ………..
? (eē + 1 – 2e)/2e = 1/2! + 1/4! + ………..
? (e – 1)ē /2e = 1/2! + 1/4! + ………..
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
Answer
Answer: (b) 45
Hint:
here it is given that T3 = 4.
? arē = 4
Now product of first five terms = a.ar.arē.arģ.ar4
= a5r10
= (ar2)5
= 45
Question 8:
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
Answer
Answer: (c) 1
Hint:
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 9.
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
Answer
Answer: (c) 6
Hint:
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 10.
If the sum of the roots of the quadratic equation axē + 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.
Answer
Answer: (c) H.P.
Hint:
Given, equation is
axē + 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/pē + 1/qē
? p + q = (pē + qē)/(pē Ũqē)
? p + q = {(p + q)ē – 2pq}/(pq)ē
? -b/a = {(-b/a)ē – 2c/a}/(c/a)ē
? (-b/a)Ũ(c/a)ē = {bē/aē – 2c/a}
? -bcē/aģ = {bē – 2ca}/aē
? -bcē/a = bē – 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 11.
If a, b, c are in G.P., then the equations axē + 2bx + c = 0 and dxē + 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
Answer
Answer: (a) AP
Hint:
Given a, b, c are in GP
? bē = ac
? bē – ac = 0
So, axē + 2bx + c = 0 have equal roots.
Now D = 4bē – 4ac
and the root is -2b/2a = -b/a
So -b/a is the common root.
Now,
dxē + 2ex + f = 0
? d(-b/a)ē + 2eŨ(-b/a) + f = 0
? db2 /aē – 2be/a + f = 0
? dŨac /aē – 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/bē
? d/a + f/c = 2e/b
? d/a, e/b, f/c are in AP
Question 12.
If a, b, c are in AP then
(a) b = a + c
(b) 2b = a + c
(c) bē = a + c
(d) 2bē = a + c
Answer
Answer: (b) 2b = a + c
Hint:
Given, a, b, c are in AP
? b – a = c – b
? b + b = a + c
? 2b = a + c
Question 13.
The 35th partial sum of the arithmetic sequence with terms an = n/2 + 1
(a) 240
(b) 280
(c) 330
(d) 350
Answer
Answer: (d) 350
Hint:
The 35th partial sum of this sequence is the sum of the first thirty-five terms.
The first few terms of the sequence are:
a1 = 1/2 + 1 = 3/2
a2 = 2/2 + 1 = 2
a3 = 3/2 + 1 = 5/2
Here common difference d = 2 – 3/2 = 1/2
Now, a35 = a1 + (35 – 1)d = 3/2 + 34 Ũ(1/2) = 17/2
Now, the sum = (35/2) Ũ (3/2 + 37/2)
= (35/2) Ũ (40/2)
= (35/2) Ũ 20
= 35 Ũ 10
= 350
Question 14.
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
Answer
Answer: (c) 6
Hint:
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 15.
The first term of a GP is 1. The sum of the third term and fifth term is 90. The common ratio of GP is
(a) 1
(b) 2
(c) 3
(d) 4
Answer
Answer: (c) 3
Hint:
Let first term of the GP is a and common ratio is r.
3rd term = arē
5th term = ar4
Now
? arē + ar4 = 90
? a(rē + r4) = 90
? rē + r4 = 90
? rē Ũ(rē + 1) = 90
? rē(rē + 1) = 3ē Ũ(3ē + 1)
? r = 3
So the common ratio is 3
Question 16.
If 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
(a) 10
(b) 12
(c) 11
(d) 13
Answer
Answer: (c) 11
Hint:
Given,
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, ….
? (2n/2)Ũ{2Ũ2 + (2n-1)3} = (n/2)Ũ{2Ũ57 + (n-1)2}
? nŨ{4 + 6n – 3} = (n/2)Ũ{114 + 2n – 2}
? 6n + 1 = {2n + 112}/2
? 6n + 1 = n + 56
? 6n – n = 56 – 1
? 5n = 55
? n = 55/5
? n = 11
Question 17.
If a is the A.M. of b and c and G1 and G2 are two GM between them then the sum of their cubes is
(a) abc
(b) 2abc
(c) 3abc
(d) 4abc
Answer
Answer: (b) 2abc
Hint:
Given, a is the A.M. of b and c
? a = (b + c)
? 2a = b + c ………… 1
Again, given G1 and G1 are two GM between b and c,
? b, G1, G2, c are in the GP having common ration r, then
? r = (c/b)1/(2+1) = (c/b)1/3
Now,
G1 = br = bŨ(c/b)1/3
and G1 = br = bŨ(c/b)2/3
Now,
(G1)ģ + (G2)3 = bģ Ũ(c/b) + bģ Ũ(c/b)ē
? (G1)ģ + (G2)ģ = bģ Ũ(c/b)Ũ( 1 + c/b)
? (G1)ģ + (G2)ģ = bģ Ũ(c/b)Ũ( b + c)/b
? (G1)ģ + (G2)ģ = bē ŨcŨ( b + c)/b
? (G1)ģ + (G2)ģ = bē ŨcŨ( b + c)/b ………….. 2
From equation 1
2a = b + c
? 2a/b = (b + c)/b
Put value of(b + c)/b in eqaution 2, we get
(G1)ģ + (G2)ģ = bē Ũ c Ũ (2a/b)
? (G1)ģ + (G2)ģ = b Ũ c Ũ 2a
? (G1)ģ + (G2)ģ = 2abc
Question 18.
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
Answer
Answer: (a) n/(n+1)
Hint:
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 19.
If 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
(a) 228
(b) 74
(c) 740
(d) 1090
Answer
Answer: (c) 740
Hint:
Let a is the first term and d is the common difference of AP
Given the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term
? a + 2d = 7 ………….. 1
and
3(a + 2d) + 2 = a + 6d
? 3Ũ7 + 2 = a + 6d
? 21 + 2 = a + 6d
? a + 6d = 23 ………….. 2
From equation 1 – 2, we get
4d = 16
? d = 16/4
? d = 4
From equation 1, we get
a + 2Ũ4 = 7
? a + 8 = 7
? a = -1
Now, the sum of its first 20 terms
= (20/2)Ũ{2Ũ(-1) + (20-1)Ũ4}
= 10Ũ{-2 + 19Ũ4)}
= 10Ũ{-2 + 76)}
= 10 Ũ 74
= 740
Question 20.
If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then
(a) a, b, c are in AP
(b) aē, bē, cē are in AP
(c) 1/1, 1/b, 1/c are in AP
(d) None of these
Answer
Answer: (b) aē, bē, cē are in AP
Hint:
Given, 1/(b + c), 1/(c + a), 1/(a + b)
? 2/(c + a) = 1/(b + c) + 1/(a + b)
? 2bē = aē + cē
? aē, bē, cē are in AP
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