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

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, 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.

(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

= (−171)

Therefore, λ = −171

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

(a) 300

(b) 175

(c) 225

(d) 150

Answer: (d)

Solution: a_n is a positive term of GP.

Let G.P. be a, ar, ar²,…..

200 = ar² + ar⁴ + …….+ ar²⁰⁰

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

Also,

100 = a₂ + a₄ + …+ a₂₀₀

⇒ 100 = ar + ar³ + ……+ ar¹⁹⁹

100 = [ar(r²⁰⁰ – 1)]/[r²-1] …..(2)

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

And,

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

⇒ ar + ar² + ar³ + ….+ ar²⁰⁰ = 300

⇒ r(a + ar + ar² + …..+ ar¹⁹⁹) = 300

⇒ 2(a₁ + a₂ + a₃ + …….+ a₂₀₀) = 300

Question 4: If x, y, z are in G.P. and a^x = b^y = c^z, find the relation between a and b.

Solution:

x, y, z are in G.P., then y² = x z..(i)

Given a^x = b^y = c^z

Let x log a = y log b = z log c = k

x =k/log a

y = k/log b

z = k/log c

Put x, y , z in (i)

k²/(log b)² = k²/(log a)(log c)

log a /log b =log b/log c

⇒ log_ba = log_cb

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, the first term = b

common ratio = r

For infinite series: −1 < r < 1

And, S= b/(1−r) = 5

⇒ b = 5(1 − r)

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

0 < 1−r < 2

⇒ 0 < b <10

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

Question 7: If the 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 × ar × ar² × ar³ × ar⁴ × ar⁵ × ar⁶ × ar⁷ × ar⁸

= (ar⁴)⁹ = 2⁹ = 512

Question 8: Let a, b, and 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: (b)

Solution:

Here b/a = c/b

⇒ c = b²/a and

⇒ a – 2b + c = 6

Using the value of c, we have

a – 2b + b²/a = 6

The above quadratic equation can 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

So [a² + a – 14]/[a + 1] = (36 + 6 – 14)/7

= 28/7

= 4

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, and 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 the 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 – r)²

⇒ 1 + r = 4 – 4r

⇒ r = 3/5

Also Read

Sequence Series JEE Advanced Previous Year Questions With Solutions

Geometric Progression and Its Properties