# CBSEtips.in

## Sunday, 21 February 2021

### CBSE Class 11 Maths - MCQ and Online Tests - Unit 9 - Sequences and Series

#### CBSE Class 11 Maths  – MCQ and Online Tests – Unit 9 – Sequences and Series

Every year CBSE schools conducts Annual Assessment exams for 6,7,8,9,11th standards. These exams are very competitive to all the students. So our website provides online tests for all the 6,7,8,9,11th standard’s subjects. These tests are also very effective and useful for those who preparing for any competitive exams like Olympiad etc. It can boost their preparation level and confidence level by attempting these chapter wise online tests.

These online tests are based on latest CBSE syllabus. While attempting these our students can identify the weak lessons and continuously practice those lessons for attaining high marks. It also helps to revise the NCERT textbooks thoroughly.

#### 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

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

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: (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

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: (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: (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

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

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

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.

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

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: (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

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

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

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

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

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

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

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: (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

Share: