Description

This mock test of Test: Sequences And Series (CBSE Level) - 1 for JEE helps you for every JEE entrance exam.
This contains 25 Multiple Choice Questions for JEE Test: Sequences And Series (CBSE Level) - 1 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Sequences And Series (CBSE Level) - 1 quiz give you a good mix of easy questions and tough questions. JEE
students definitely take this Test: Sequences And Series (CBSE Level) - 1 exercise for a better result in the exam. You can find other Test: Sequences And Series (CBSE Level) - 1 extra questions,
long questions & short questions for JEE on EduRev as well by searching above.

QUESTION: 1

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

Solution:

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

QUESTION: 2

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

Solution:

The given sequence is based on the following pattern:

∴ Required number = 30.

QUESTION: 3

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

Solution:

QUESTION: 4

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

Solution:

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

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

Solution:

QUESTION: 6

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

Solution:

QUESTION: 7

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

Solution:

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.

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

Solution:

S_{p} = Sq

⇒ p/2(2a+(p−1)d) = q/2(2a+(q−1)d)

⇒ p(2a+(p−1)d) = q(2a+(q−1)d)

⇒ 2ap + p^{2}d − pd = 2aq + q^{2}d − 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

⇒ S_{p+q} = 0

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

Solution:

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

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

Solution:

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(p^{2}+q^{2}-p-q+pq)/pq - (p+q-1) 2 (p+q)/pq}

i.e. Sp+q = (p+q)/pq { p^{2}+q^{2}-p-q+pq-p^{2}-pq-qp-q^{2}+p+q}

So, Sp+q = -(p+q)

QUESTION: 11

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

Solution:

QUESTION: 12

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

Solution:

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

QUESTION: 13

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

Solution:

QUESTION: 14

The number of numbers between n and n^{2} which are divisible by n is

Solution:

Between n & n^{2}, numbers divisible by n are:

2n, 3n, 4n, ….... (n – 1)n

No. of numbers = (n – 1) – 2 + 1 = n – 2

QUESTION: 15

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

Solution:

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

QUESTION: 16

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

Solution:

QUESTION: 17

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

Solution:

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

Solution:

QUESTION: 19

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

Solution:

QUESTION: 20

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

Solution:

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

Solution:

QUESTION: 22

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

Solution:

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

⇒ab+cb+ad+cd = 4ad

⇒ab+bc+cd = 3ad

QUESTION: 23

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

Solution:

The 2-digit number which when divided by 3 gives remainder 1 are: 10, 13, 16, ...97

Here a = 10, d = 13 - 10 = 3

t_{n} = 97

nth term of an AP is t_{n} = 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

QUESTION: 24

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

Solution:

100÷7= 14 2/7 , 7×15=105

1000÷7=142 6/7 , 7×142=994

15th to 142 th multiples of 7 are there in between 100 and 1000 .

No of them are142–14= 128

so 128 is the ans

QUESTION: 25

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

Solution:

### Notes : Sequences & Series

Doc | 13 Pages

### Introduction to series - Sequences & Series

Video | 02:43 min

### Sequences and Series (part 1) - Calculus, Mathematics

Video | 09:49 min

### Sequences n series

Doc | 26 Pages

- Test: Sequences And Series (CBSE Level) - 1
Test | 25 questions | 25 min

- Test: Sequences And Series (CBSE Level) - 2
Test | 25 questions | 25 min

- Test: Sequences And Series (Competition Level) - 1
Test | 22 questions | 44 min

- Test: Sequences And Series - 1
Test | 15 questions | 30 min

- Test: Sequences And Series - 1
Test | 10 questions | 10 min