Test: Previous Year Questions: Sequence & Series- 1


40 Questions MCQ Test Mathematics For JEE | Test: Previous Year Questions: Sequence & Series- 1


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QUESTION: 1

The sum of the series 13 – 23 + 33 – ......+ 93 =

[AIEEE- 2002]

Solution:

(13 + 33 + .....+ 93) - (23 + 43 + 63+ 83)
⇒ (13 + 23 + 33 + ... + 93) - 2.(23 + 43+63 + 83
 - 2.23 (1 + 23 + 33 +43)
⇒ (45)2 - 16 
⇒ (45)2 - (40)2
⇒ 85 (5) ⇒ 425 

QUESTION: 2

If the sum of an infinite GP is 20 and sum of their square is 100 then  common ratio will be =

[AIEEE- 2002]

Solution:

ANSWER :- c

Solution :- We consider the infinite GP 

a,ar,ar^2,..,ar^(n−1),....

We know that, for this GP, the sum of its infinite no. of terms is

S∞ = a/(1−r).

∴ a/(1−r) = 20.........................(1).

The infinite series of which, the terms are the squares of the

terms of the first GP is, a^2+a^2r^2+a^2r^4+... + a^2r^(2n−2)+....

We notice that this is also a Geom. Series, of which the

first term is a^2 and the common ratio r^2.

Hence, the sum of its infinite no. of terms is given by,

S∞ = a^2(1−r^2).

∴ a^2/(1−r^2)= 100........................(2).

(1)÷(2) ⇒(1+r)/a= 1/5......................(3).

Then, (1) × (3) gives ,(1+r)/(1−r) = 4.

⇒ r = 3/5, is the desired common ratio!

QUESTION: 3

If the third term of an A.P. is 7 and its 7th term is 2 more than three times of its 3rd term, then sum of its first 20 terms is-

[AIEEE- 2002]

Solution:

Let AP be a, a + d, a + 2d, .....
∴ T3 = a + 2d = 7 ....(i)
& T7 = a + 6d = 2 + 3 (7) = 23 .....(ii)
On solving (i) & (ii), we get
d = 4 and a = (–1)

QUESTION: 4

If x1, x2, x3 and y1, y2, y3 are both in G.P. with the same common ratio, then the points (x1, y1), (x2, y2) and (x3, y3)

[AIEEE- 2003]

Solution:

x1, x2, x3 and y1, y2, y3 are in GP with same common ratio,
∴ (x1,y1) ⇒ p(x1,y1)
(x2,y2) ⇒ Q(x1r,y1r)
(x3,y3) ⇒ R(x1r2,y1​r2)

∴ P, Q, R are collivear

QUESTION: 5

Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation-

[AIEEE- 2004]

Solution:

Let the numbers be a & b,

a + b = 18 & ab = 16
∴ Q.E. with roots a & b is
x2- (a + b)x + ab = 0
⇒ x2 - 18x + 16 = 0 

QUESTION: 6

Let Tr be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, m ≠ n, Tm = 1/n  and Tn = 1/m, then a - d equals-

[AIEEE- 2004]

Solution:

Given that, Tm = 1/n
⇒ a + (m - 1) d = 1/n ......(i)
and Tn = 1/m
⇒ a + (n - 1)d = 1/m  ......(ii)
On solving Eqs. (i) and (ii), we get
a = d = 1/mn
⇒ a - d = 0

QUESTION: 7

The sum of the first n terms of the series 12 + 2. 22 + 32 + 2.42 + 52 + 2.62 +..... is  when n is even. When n is odd the sum is-

[AIEEE- 2004]

Solution:

Given that, the sum of n terms of given series is if n is even.
Let n is odd ie, n = 2m + 1
Then, S2m+1 = S2m + (2m + 1)th term
 

QUESTION: 8

 where a, b, c are in A.P. and I a I < 1, l b I < 1, I c I < 1 then x, y, z are in -

[AIEEE- 2005]

Solution:

Given that,

 .........(ii)

Now, a,b,c are in AP.

⇒ -a, -b, -c are in AP.
⇒ 1 - a, 1 - b, 1 - c are  also in AP.
 are in AP.
⇒ x, y,z are in HP.

QUESTION: 9

If in a ΔABC, the altitudes from the vertices A, B, C on opposite sides are in H.P., then sin A, sin B, sin C are in

[AIEEE- 2005]

Solution:




⇒ a,b,c are in AP
⇒ sin A, sin B, sin C are in AP

QUESTION: 10

Let a1, a2, a3, ..... be terms of an A.P. 

[AIEEE- 2006]

Solution:



Where d be a common difference of an AP.


QUESTION: 11

If a1, a2, ..... an are in H.P., then the expression a1a2 + a2a3 +....+ an –1an is equal to –

[AIEEE- 2006]

Solution:

Let d be the common difference of AP.


On adding all of these, we get

On putting the value of d in Eq. (i), we get

QUESTION: 12

In a geometric progressionconsisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression equals-

Solution:

Since, each term is equal to the sum of two preceeding terms.

QUESTION: 13

The sum to infinity of the series 

[AIEEE 2009]

Solution:


On subtracting Eq. (ii) from Eq. (i), we get

QUESTION: 14

A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If a1 = a2 = .... = a10 = 150 and a10, a11 .... are in an AP with common diferencec –2, then the time taken by him to count all notes is -

[AIEEE 2010]

Solution:

Number of notes that the person counts in 10 min
= 10 × 150 = 1500
Since, a10, a11, a12, ..... are in AP with common difference –2.
⇒ Let n be the time taken to count remaining 3000 notes, then

⇒ n2 - 149n + 3000 = 0
⇒ (n - 24) (n - 125) = 0
⇒ n = 24, 125
The, the total time taken by the person to count all notes
= 10 + 24 = 34 min.

QUESTION: 15

Statement 1 : The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + ...... ....+(361 + 380 + 400) is 8000.
Statement 2 : for any natural number n.

 [AIEEE- 2012]

Solution:

Statement - I : 1+(1+2+4)+(4+ 6+9)+(9+12+16) +..... + (361 + 380 +400)
⇒ 1 + (23- 13) + (33 - 23) + (43- 33) + ....+ (20- 193
⇒ (20)3 ⇒ 8000 
Statement - II : is true and statement - II is correct explanation of Statement - I

QUESTION: 16

If 100 times the 100th term of an AP with non zero common difference equals the 50 times its 50 th term, then the 150th term of this AP is:

[AIEEE 2012]

Solution:

Let the AP be a, a + d, a + 2d, ..... where d ≠ 0
Now, 100 (T100) = 50 (T50)
⇒100 (a + 99d) = 50 (a + 49d)
⇒ 50 a = –2450 d ⇒ a = (–149 d)
Now, T150 = a + 149 d = 0

QUESTION: 17

Consider an infinite geometric series with first term a and common ratio r. If the sum is 4 and the second term is 3/4, then

[JEE 2000,]

Solution:


⇒ 3 = 4r(4-4r)
⇒ 16r2 -16r+3 = 0 ⇒ (4r-3)(4r-1) = 0 
∴ r = 3/4 & a = 1 or r = 1/4 & a = 3 

QUESTION: 18

If a, b, c, d are positive real numbers such that a + b + c + d = 2, then M = (a + b) (c + d) satisifes the relation :

Solution:

a + b + c + d = 2
 


M is greater than or equal to 0 & less than or equal to 1 or 0 ≤ M ≤ 1

QUESTION: 19

Given that α,γ are roots of the equation, Ax2–4x+1 = 0 and β, δ the roots of the equation, Bx2 – 6x + 1 = 0, find values of A and B, such that

 [REE 2000, 5]

Solution:



γ satisfy equation (1)

β satisfy equation (2)

QUESTION: 20

Le α, β be the roots of x2 – x + p = 0 and γ, δ the roots of x2 – 4x + q = 0. If α, β, γ, δ are in G.P., then the integral values of p and q respectively, are

Solution:

α, β, γ, δ in GP.
Let α = a, β = ar, γ = ar2, δ = ar3
a+ar = 1  &  ar2+ar= 4
a(1+r) = 1   ar2 (1 + 1) = 4
r2 = 4 = r = ±2 



satisfy given equations but p & q not integer for r = 2
⇒ r ≠ 2
∴ r = -2a = -1 ⇒ α = -1,β = 2,γ = -4,δ = 8
(-1)2-(-1)+P = 0 ⇒ p =-2
8,(-4)2-4(-4)+q = 0 ⇒ q = -32 

QUESTION: 21

Suppose a, b, c are in A.P. and a2, b2, c2 are in G.P. if a < b < c and a + b + c = 3/2, then the value of a is

[JEE 2002 ]

Solution:

2b =  a + c & (b2)2 = a2c2 & a < b < c

& a + b + c = 3/2 & b2 = + ac
⇒ 3b = 3/2

roots of eqaution 
Case - I 
⇒ 4x2 - 4x + 1 = 0 = (2x- 1)2 = 0 
⇒ a= c (but a < b < c) ∴ reject
Case -II 

QUESTION: 22

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

Solution:

2 + 5 + 8 + .....+ T2n = 57 + 59 + 61 + ....+ Tn


⇒ n [4 + 6n – 3] = n [57 + (n – 1)]
⇒  6n + 1 = n + 56
⇒  5n = 55
⇒ n = 11

QUESTION: 23

Let the positive numbers a, b, c, d be in A.P. Then abc, abd, acd, bcd are

 [JEE 2001, (Scr.)]

Solution:



QUESTION: 24

The first term of an infinite geometric progression is x and its sum is 5. Then  

 [JEE 2004 (Scr.)]

Solution:


QUESTION: 25

In the quadratic equati on ax2 + bx + c = 0, If Δ = b2 – 4ac and α + β, α2 + β2, α3+ β3, are in G.P. where α , β are the roots of ax2 + bx + c = 0, then    

[JEE 2005 (Scr.)]

Solution:


Δ = b2 — 4ac, quadratic ⇒ a ≠ 0
α + β, α2 + β2, α3 + β3 in GP. 
2 + β2)= (α + β) (α3 + β3)
⇒ α4 + β4 + 2α2β= α4 + β + αβ(α2 + β2)
⇒ αβ(α2 + β2 - 2αβ) = 0
⇒ αβ (α - β)2 = 0

⇒ cΔ = 0

QUESTION: 26

Let Vr d enote the sum of first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (2r – 1). Let Tr = V r + 1 – Vr – 2 and Qr = Tr + 1 – Tr for r = 1, 2,.....
The sum V1 + V2 + ....... + Vn  is

Solution:



QUESTION: 27

Let Vr denote the sum of first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (2r – 1). Let Tr = V r + 1 – Vr – 2 and Qr = Tr + 1 – Tr for r = 1, 2,.....
Tr is always

Solution:

Vr+1 - Vr



which is composite no.

QUESTION: 28

Le t Vr d enote the sum of first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (2r – 1). Let Tr = V r + 1 – Vr – 2 and Qr = Tr + 1 – Tr for r = 1, 2,.....
Which one of the following is a correct statement ?

Solution:

T= 3r2 + 2r-1
Tr+1 = 3(r+ 1)2 + 2(r + 1)-1
Qr = Tr+1 - Tr = 3(2r + 1)+ 2 (1)
Qr =  6r + 5
Qr+1 = 6(r + 1)+ 5
common difference = Qr+ 1 - Qr = 6

QUESTION: 29

Let A1, G1, H1 denote the arithmetic, geometric and harmoni c means, respectively, of two distinct positive numbers. For n ≥ 2, Let An – 1 and Hn – 1 have arithmetic, geometric and harmonic means as An, Gn, Hn respectively
Which one of the following statements is correct ?

Solution:





QUESTION: 30

G1,G2,…,Gn are said to be n geometric means between a and b if a,G1,…Gn,b is

Solution:
QUESTION: 31

Let A1, G1, H1 denote the arithmetic, geometric and harmonic means, respectively, of two distinct positive numbers. For n ≥ 2, let An−1 and Hn−1 has arithmetic, geometric and harmonic means as An, Gn, Hn respectively.
Which of the following statements is correct?

Solution:

A3 is A.M. of A1 an H1 and A1 > H1 ⇒ A1 > A2 > H1
A3 is A.M. of A2 an H2 and A2 > H2 ⇒ A2 > A3 > H2
∴ A1 > A2 > A3 > ...

QUESTION: 32

Let A1, G1, H1 denote the arithmetic, geometric and harmoni c means, respectively, of two distinct positive numbers. For n ≥ 2, Let An – 1 and Hn – 1 have arithmetic, geometric and harmonic means as An, Gn, Hn respectively

Solution:

As above A1 > H2 > H1 & A2 > H3 > H2 
∴ H1 > H2 < H3 < ......

QUESTION: 33

A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and the circumcircle of the triangle PQR at the point T. If S is not the centre of the circumcircle, then

[JEE 2008, 4]

Solution:

 (PS) × (ST) = (QS) × (SR)


 ......(i)

 ...(ii)
from (i) & (ii)

QUESTION: 34

Supose four distinct positive numbers a1, a2, a3, a4 are i n G. P. Let b1 = a1, b2 = b1 + a2, b3 = b2 + a3 and b4 = b3 + a4.  
Statement–1 : The numbers b1, b2, b3, b4 are neither in A.P. nor in G.P.
Statement–2 : The numbers b1, b2, b3, b4 are in H.P.

[JEE 2008]

Solution:

b1 = a1
b2 = a1 + a2
b3 = a1 + a2 + a3
b4 = a1 + a2 + a3 + a4
Hence b1, b2, b3, b4 are neither in A.P. nor in G.P. nor in H.P.

QUESTION: 35

If the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is  

[JEE 2009]

Solution:









QUESTION: 36

Let Sk, K = 1, 2, ...., 100 denote the sum of the infinite geometric series whose first term is and the common ratio is 1/k. Then the value of 

Solution:


but k ≠ 1







QUESTION: 37

Let a1, a2, a3, ....., a11 be real numbers satisfying a1 = 15, 27 – 2a2 > 0 and ak = 2ak-1 – ak-2 for k = 3, 4, ...., 11

If then yhe value of  is equal to 

 [JEE 2010]

Solution:

= ak = 2ak-1 - ak-2
⇒ a1, a2, a3 ...., a11 are in A.P. with common difference d.
a1 = 15,a2 = 15+d, a3 = 15+2d ..... a11 = 15+ 10d 


⇒ 35d2 + 150d + 225 = 90
⇒ 35d2 + 150d + 135 = 0
⇒ 7d2 + 30d + 27 = 0



QUESTION: 38

The minim um value of the s um of re al numb ers a –5, a–4, 3a–3, 1, a8 and a10 with a > 0 is

[JEE 2011]

Solution:

Using AM > GM

⇒ Sum = 8

QUESTION: 39

Let a1, a2, a3, ..., a100 be an arithmetic progression with a1 = 3 and   For any integer n with 1≤ n ≤ 20. let m=5n. If Sm/Sdoes not depend on n, then a2 is

 [JEE 2011]

Solution:



QUESTION: 40

Let a1, a2, a3, ..... be in harmonic progression with a1 = 5 and a20 = 25. The least positive integer n for which an < 0 is 

[JEE 2012]

Solution: