JEE Main Geometric Progression Previous Year Questions With Solutions

Past year JEE Main solved problems on geometric progression are available here. A geometric sequence is a sequence of numbers in which the ratio of consecutive terms is always constant. This constant term is called the common ratio, which can be represented as “r”. The value of the common ratio determines whether the G.P. is decreasing, increasing, positive or negative. In this section, students can get a list of solved questions asked in previous year JEE Main exams.

Download Geometric Progression Previous Year Solved Questions PDF

Important Points

Definition:

A geometric progression is a sequence where the succeeding term is ‘r’ times the preceding term.

Representation: a, ar, ar2, ……, arn-1, arn

Where a = first term and r = common ratio

Formula to find common ratio:

r = an/an-1

General term/nth term:

an = arn-1

Sum of n terms:

Sn = a(1-rn)/(1-r) if r < 1 and

Sn = a(rn – 1)/(r-1) if r > 1

JEE Main Past Year Questions With Solutions on G.P.

Question 1: a, b, c are in G.P., a + b + c = bx, then x can not be

(a) 2                     (b) -2                      (c) 3                  ( d) 4

Answer: (a)

Solution:

Let the terms of G.P. be a/r, a, ar

Therefore, a/r + a + ar = ax

=> x = r + 1/r + 1

But r + 1/r ≥ 2 or r + 1/r ≤ -2 [By A.M and G.M. inequality]

Therefore,

But x – 1 ≥ 2 or x – 1 ≤ -2

=> x ≥ 3 or x ≤ -1

Question 2: Let a1, a2, a3, …. be a G.P. such that a1 < 0, a1 + a2 = 4 and a3 + a4 = 16. If i=19ai=4λ\sum_{i=1}^9a_i = 4 \lambda, then λ is equal to

(a) 171                     (b) 511/3                     (c) -171                    (d) -513

Answer: (c)

Solution:

a1 + a2 = 4

=> a + ar = 4

=> a(1 + r) = 4

And, a3 + a4 = 16

=> ar2 + ar3 = 16

=> ar2(1 + r) = 16

=> 4r2 = 16

=> r = ±2

If r = 2, a = 4/3; which is not possible as a1 < 0

If r = −2, a = −4

i=19ai=a(r91r1=4((2)91)3=43(5121)\sum_{i=1}^9a_i = \frac{a(r^9-1}{r-1} = \frac{-4((-2)^9 – 1)}{-3} = \frac{4}{3}(-512-1)

= 4(−171)

Therefore, λ = −171

Question 3: Let an be the nth term of a G.P. of positive terms.

If n=1100a2n+1=200  and  n=1100a2n=100  then  n=1200an\sum_{n=1}^{100}a_{2n+1} = 200 \; and \; \sum_{n=1}^{100}a_{2n} = 100 \; then \; \sum_{n=1}^{200}a_{n} is equal to:

(a) 300               (b) 175                  (c) 225                   (d) 150

Answer: (d)

Solution: an is a positive term of GP.

Let G.P. be a, ar, ar2,…..

n=1100a2n+1=a3+a5+.+a201\sum_{n=1}^{100} a_{2n+1} = a_3 + a_5 + …….+a_{201}

200 = ar2 + ar4 + …….+ ar201

=> 200 = [ar2(r200−1)]/[r2−1] ….. . . (1)

Also, n=1100a2n=100\sum_{n=1}^{100} a_{2n} = 100

100 = a2 + a4 + …+ a200

=> 100 = ar + ar3 + ……+ ar199

100 = [ar(r200 – 1)]/[r2-1] …..(2)

From (1) and (2), we have, r = 2

And, n=1100a2n+1+n=1100a2n=300\sum_{n=1}^{100} a_{2n+1} + \sum_{n=1}^{100} a_{2n} = 300

=> a2 + a3 + a4 + …….+ a200 + a201 = 300

=> ar + ar2 + ar3 + ….+ ar200 = 300

=> r(a + ar + ar2 + …..+ ar199) = 300

=> 2(a1 + a2 + a3 + …….+ a200) = 300

n=1200an=150\sum_{n=1}^{200} a_{n} = 150

Question 4: If x, y, z are in G.P. and ax = by = cz, then find the relation between a and b.

Solution:

x, y, z are in G.P., then y2 = x * z

Now ax = by = cz = m

⇒ x loge a = y loge b = z loge c = loge m

⇒ x = loga m, y = logb m, z = logc m

Again, as x, y, z are in G.P., so

(p − r)2 = (p + r)2 −4pr

= 16K2 − 16K2 = 0

⇒ logbm / logam = logcm / logbm

⇒ logba = logcb

Question 5: Consider an infinite G.P. with first term a and common ratio r, its sum is 4 and the second term is 3/4, then find a and r.

Solution:

Here a / [1 − r] = 4 and ar = 3/4.

Dividing these, r (1 − r) = 3 / 16 or 16r2 − 16r + 3 = 0 or

(4r − 3) (4r − 1) = 0

or r = 1/4,  3/4 and

a=3, 1

So (a, r) = (3, 1/4), (1, 3/4).

Question 6: If b is the first term of an infinite G.P. whose sum is 5, then b lies in the interval,

(a) (-∞, -10)                (b) (10, ∞)              (c) (0, 10)                    (d) (10, 0)

Answer: (c)

Solution: Here, first term = b

common difference = d

For infinite series: −1 < d < 1

And, S= b/(1−d) = 5

=> b = 5(1 − d)

Since |d| < 1 for an infinite geometric series to converge, we have −1 < d < 1, This implies

0 < 1−d < 2

=> 0 < b <10

So, b lies in the interval (0, 10).

Question 7: If fifth term of G.P. is 2. Find the product of its nine terms.

Solution:

Given, a5 = 2

ar4 = 2

Product of nine terms = a x ar x ar2 x ar3 x ar4 x ar5 x ar6 x ar7 x ar8

= (ar4)9 = 29 = 512

Question 8: Let a, b, c be positive integers such that b/a is an integer. If a, b, c are in G.P. and the arithmetic mean of a, b, c is b+2, then the value of [a2+a-14]/[a+1] is

(a) 3                         (b)  4                                (c) 6                         (d) 5

Answer: (c)

Solution:

Here b/a = c/b

=> c = b2/a and

[a+b+c]/3 = b + 2

=> a – 2b + c = 6

Using value of C, we have

a – 2b + b2/a = 6

Above quadratic equation cab be written as,

(b/a – 1) 2 = 6/a

=> (b/a – 1)2 is a perfect square

Therefore, 6/a is a perfect square only if a = 6

Question 9: If the 4th, 7th and 10th terms of G.P. be a, b, c respectively, then the relation between [a, b, c] is

(a) b = [a+c]/2

(b) a2 = bc

(c) b2 = ac

(d) c2 = ab

Answer: (c)

Solution:

Given: 4th, 7th and 10th terms of G.P. be a, b, c.

T4 = ar3 = a

T7 = ar6 = b

T10 = ar9 = c

Since a, b, c are in G.P.

=> b2 = ac; relation is true.

Verify: b2 = (ar6) 2 = a2r12 and ac = ar3 x ar9 = a2r12

Question 10: Sum of infinite number of terms of G.P. is 20 and sum of their square is 100. The common ratio of G.P. is

(a) 5 (b) 3/5 (c) 8/5 (d) 1/5

Answer: (b)

Solution:

Let G.P. be a, ar, ar2

Sum of infinite terms = 20 = a/(1-r)

=> a = 20(1 – r) ….(i)

Again, Sum of their square = a2 + a2r2 + a2r4 + …. to infinity = 100

=> a2/(1-r2) = 100

=> a2 = 100(1 – r2) ….(ii)

Using (i) and (ii), we have

100(1 – r)(1 + r) = 400(1 – r2)

=> 1 + r = 4 – 4r

=> r = 3/5

 

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