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Test: JEE Previous Year Questions- Sequence & Series- 1 - GRE MCQ


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30 Questions MCQ Test Mathematics for GRE Paper II - Test: JEE Previous Year Questions- Sequence & Series- 1

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Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 1

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

[AIEEE- 2002]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 1

(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 

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 2

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!

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Test: JEE Previous Year Questions- Sequence & Series- 1 - 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]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 3

Test: JEE Previous Year Questions- Sequence & Series- 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]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 4

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

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 5

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 

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 6

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

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 7

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
 

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 8

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.

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 9




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

Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 10

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

[AIEEE- 2006]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 10



Where d be a common difference of an AP.


Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 11

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

[AIEEE- 2006]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 11

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

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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-

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 12

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

Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 13

The sum to infinity of the series 

[AIEEE 2009]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 13


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

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 14

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.

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 15

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

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 16

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

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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,]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 17


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

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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 :

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 18

a + b + c + d = 2
 


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

Test: JEE Previous Year Questions- Sequence & Series- 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]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 19



γ satisfy equation (1)

β satisfy equation (2)

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 20

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 

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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 ]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 21

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 

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 22

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


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

Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 23

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

 [JEE 2001, (Scr.)]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 23



Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 24

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

 [JEE 2004 (Scr.)]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 24


Test: JEE Previous Year Questions- Sequence & Series- 1 - 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.)]

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 25


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

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 26



Test: JEE Previous Year Questions- Sequence & Series- 1 - 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

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 27

Vr+1 - Vr



which is composite no.

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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 ?

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 28

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

Test: JEE Previous Year Questions- Sequence & Series- 1 - 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 ?

Detailed Solution for Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 29





Test: JEE Previous Year Questions- Sequence & Series- 1 - Question 30

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

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