Test: Sequence and Series - JEE MCQ

# Test: Sequence and Series - JEE MCQ

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## 25 Questions MCQ Test Physics for JEE Main & Advanced - Test: Sequence and Series

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Test: Sequence and Series - Question 1

### The next term of the sequence 1, 2, 4, 7,11,…. is

Detailed Solution for Test: Sequence and Series - Question 1

The given series is: 1,2,4,7,11,...
Difference between second and first term = 2 - 1 = 1
Difference between third and second term = 4 - 2 = 2
Difference between fourth and third term = 7 - 4 = 3
Difference between fifth and fourth term = 11 - 7 = 4
Difference between sixth and fifth term = 16 - 11 = 5

Test: Sequence and Series - Question 2

### The next term of the sequence, 2, 6, 12, 20, …..is

Detailed Solution for Test: Sequence and Series - Question 2

The given sequence is based on the following pattern:

∴ Required number = 30.

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Test: Sequence and Series - Question 3

### The next of the series 3/2 + 5/4 + 9/8 + 17/16 .....is

Detailed Solution for Test: Sequence and Series - Question 3

Test: Sequence and Series - Question 4

Which term of the sequence 8 – 6i, 7 – 4i, 6 – 2i, ….is a real number ?

Detailed Solution for Test: Sequence and Series - Question 4

a = 8−6i
d = 7−4i−8+6i
= −1+2i
an = a+(n−1)d
a+ib = 8−6i+(n−1)(−1+2i)
a+ib = 8−6i+(−1)(n−1)+(n−1)2i
= − 6+2(n−1)=0
= 2(n−1) = 6
n = 4
an = 8−6i+(4−1)(−1+2i)
= 8−6i−3+6i = 5
4th term = 5

Test: Sequence and Series - Question 5

All the terms in A.P., whose first term is a and common difference d are squared. A different series is thus formed. This series is a

Detailed Solution for Test: Sequence and Series - Question 5

Test: Sequence and Series - Question 6

If a, 4, b are in A.P.; a, 2, b are in G.P.; then a, 1, b are in

Test: Sequence and Series - Question 7

The eleventh term of the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, ….. is

Detailed Solution for Test: Sequence and Series - Question 7

The sequence is the Fibonacci series
1+1 = 0
1+2 = 3
2+3 = 5
3+5 = 8
5+8 = 13
8+13 = 21
13+21 = 34
21+34 = 55
34+55 = 89
The 11th term will be 89.

Test: Sequence and Series - Question 8

In an A.P., sum of first p terms is equal to the sum of first q terms. Sum of its first p + q terms is

Detailed Solution for Test: Sequence and Series - Question 8

Sp = Sq
⇒  p/2(2a+(p−1)d) = q/2(2a+(q−1)d)
⇒ p(2a+(p−1)d) = q(2a+(q−1)d)
⇒ 2ap + p2d − pd = 2aq + q2d − qd
⇒ 2a(p−q) + (p+q)(p−q)d − d(p−q) = 0
⇒ (p−q)[2a + (p+q)d − d] = 0
⇒ 2a + (p+q)d − d = 0
⇒ 2a + ((p+q) − 1)d = 0
⇒ Sp+q = 0

Test: Sequence and Series - Question 9

The sum of 40 A.M.’s between two number is 120. The sum of 50 A.M.’s between them is equal to

Detailed Solution for Test: Sequence and Series - Question 9

Let A1, A2, A3, ........ , A40 be 40 A.M's between two numbers 'a' and 'b'.
Then,
a, A1, A2, A3, ........ , A40, b is an A.P. with common difference d  = (b - a)/(n + 1) = (b - a)/41
[ where n = 40]
now, A1, A2, A3, ........ , A40  = 40/2( A1 + A40)
A1, A2, A3, ........ , A40 = 40/2(a + b)
[ a, A1, A2, A3, ........ , A40, b is an Ap then ,a + b = A1 + A40]
sum of 40A.M = 120(given)
120= 20(a + b)
=> 6 = a + b ----------(1)
Again, consider B1, B2, ........ , B50  be 50 A.M.'s between two numbers a and b.
Then, a, B1, B2, ........ , B50, b will be in A.P. with common difference = ( b - a)/51
now , similarly,
B1, B2, ........ , B50 = 50/2(B1 + B2)
= 25(6) ----------------from(1)
= 150

Test: Sequence and Series - Question 10

In an A.P., sum of first p terms is q and sum of first q terms is p. Sum of its p + q terms is

Detailed Solution for Test: Sequence and Series - Question 10

Let the first term of the given AP be ‘a' and the common difference be ‘d'. Then, the sum of first ‘n' terms of the AP is given by:
Sn =  n/2 {2a+(n-1)d} …….(1)
Here, it is given that:
Sp = q and Sq = p
Using (1), we get:-
q = p/2 {2a+(p-1)d}
and p = q/2 {2a+(q-1)d}
i.e. 2a + (p-1)d = 2q/p …..(2)
and 2a + (q-1)d = 2p/q …..(3)
Subtracting (3) from (2), we get:
(p - 1 - q + 1)d = 2q/p - 2p/q
So, d = 2(q2 - p2)/pq(p-q)
i.e. d = -2(p+q)/pq
Now, substituting the value of ‘d' in eq.n (2), we get:
2a + (p-1){-2(p+q)/pq} = 2q/p
i.e. 2a= 2q/p + 2(p-1)(p+q)/pq
This gives:
a = (p2 + q2 - p - q + pq)/pq
So, we have
Sp+q = (p+q)/2 { 2(p2+q2-p-q+pq)/pq - (p+q-1) 2 (p+q)/pq}
i.e. Sp+q = (p+q)/pq { p2+q2-p-q+pq-p2-pq-qp-q2+p+q}
So, Sp+q = -(p+q)

Test: Sequence and Series - Question 11

pth term of an A.P. is q and qth term is p, its (p+ q)th term is

Test: Sequence and Series - Question 12

pth term of an H.P. is qr and qth term is pr, then rth term of the H.P. is

Detailed Solution for Test: Sequence and Series - Question 12

Given pth term of HP = qr
So pth term of AP = 1/qr
a+(p−1)d = 1/qr....(1)
and qth term of HP = pr
so qth term of AP = 1/pr
a + (q−1)d = 1/pr.....(2)
subtracting equation 1 and 2 we get,
(p−q)d = (p−q)/pqr
d = 1/pqr
Now from equation 1,
a = 1/qr − (p−1)/pqr
= (p−p+1)/pqr = 1/pqr
So rth term of AP = a+(r−1)d = 1/pqr + (r − 1)/pqr = 1/pq
So, rth term of HP = pq

Test: Sequence and Series - Question 13

The value of b for which the roots of the equation sin x = b are in A.P. is

Detailed Solution for Test: Sequence and Series - Question 13

Test: Sequence and Series - Question 14

The number of numbers between n and n2 which are divisible by n is

Detailed Solution for Test: Sequence and Series - Question 14

Between n & n2, numbers divisible by n are:
2n, 3n, 4n, ….... (n – 1)n
No. of numbers = (n – 1) – 2 + 1 = n – 2

Test: Sequence and Series - Question 15

The number of terms common to the Arithmetic progressions 3, 7, 11, …., 407 and 2, 9, 16, …., 709 is

Detailed Solution for Test: Sequence and Series - Question 15

First A.P′s sequence is 3,7,11,....,407
General term will be 4k+3    k≤101
Second A.P's sequence is 2,9,16,....,709
General term will be 7p+2    p≤101
The common terms will be 51,79,...,28m + 51
28m + 51 ≤ 407
⟹ 28k ≤ 356
⟹ k ≤ 12.71
And adding the 2 starting number count of the 2 A.P′s i.e. 3 & 2
Number of common terms will be 12 + 2 = 14

Test: Sequence and Series - Question 16

If a, b, c are in A. P. as well as in G.P.; then

Test: Sequence and Series - Question 17

The 20th term of the series 2×4+4×6+6×8+... is

Detailed Solution for Test: Sequence and Series - Question 17

Test: Sequence and Series - Question 18

The ratio of first to the last of n A.m.’s between 5 and 25 is 1 : 4. The value of n is

Detailed Solution for Test: Sequence and Series - Question 18

Test: Sequence and Series - Question 19

The next term of the sequence 1, 3, 6, 10, …. Is

Test: Sequence and Series - Question 20

If A, G, H denote respectively the A.M., G.M. and H.M. between two unequal positive quantities then

Test: Sequence and Series - Question 21

If A, G and H denote respectively, the A.M., G.M. and H.M. between two positive numbers a and b, then A - G is equal to

Detailed Solution for Test: Sequence and Series - Question 21

take lcm
a+b-2√ab
2
(√a - √b )2
2

Test: Sequence and Series - Question 22

If a, b, c, d are in H.P., then ab + bc + cd is

Detailed Solution for Test: Sequence and Series - Question 22

Since a,b,c are in H.P, so b = 2ac/(a+c).
Also, b,c,d are in H.P, so c = 2bd/(b+d).
Therefore, (a+c)(b+d) = 2ac/b × 2bd/c

Test: Sequence and Series - Question 23

The sum of all 2-digited numbers which leave remainder 1 when divided by 3 is

Detailed Solution for Test: Sequence and Series - Question 23

The 2-digit number which when divided by 3 gives remainder 1 are: 10, 13, 16, ...97
Here a = 10, d = 13 - 10 = 3
tn = 97
nth term of an AP is tn = a + (n – 1)d
97 = 10 + (n – 1)3
⇒ 97 = 10 + 3n – 3
⇒ 97 = 7 + 3n
⇒ 3n = 97 – 7 = 90
∴ n = 90/3 = 30
Recall sum of n terms of AP,
= 15[20 + 87] = 15 × 107 = 1605

Test: Sequence and Series - Question 24

The number of numbers between 105 and 1000 which are divisible by 7 is

Detailed Solution for Test: Sequence and Series - Question 24

Clearly, the numbers between 105 and 1000 which are divisible by 7 are 112,119,126,...,994.

This is an AP with first term a=112, common difference d=7 and last term  l = 994.
Let there be n terms in this AP. Then,
an​ = 994

⇒ a + (n − 1)d

994 = 112 + (n − 1) × 7

994 = 105 + 7n

∴ 7n = 889

∴ n = 127

Test: Sequence and Series - Question 25

The next term of the sequence 1, 5, 14, 30, 55, …… is

Detailed Solution for Test: Sequence and Series - Question 25

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