Geometric Progression (G.P.) — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Geometric Progression (G.P.) MCQs with step-by-step solutions (20 questions). Part of Sequences and Series. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Geometric Progression (G.P.) · easy · numerical
The $6$-th term of the G.P. $3, 6, 12, \dots$ is:
A. $96$ ✓ Correct
B. $192$
C. $48$
D. $64$
Solution: $T_6 = a r^5 = 3 \cdot 2^5 = 96$.
Q2 — Geometric Progression (G.P.) · medium · numerical
The sum of the first $6$ terms of the G.P. $2, 6, 18, \dots$ is:
A. $730$
B. $726$
C. $364$
D. $728$ ✓ Correct
Solution: $S_6 = \dfrac{2(3^6 - 1)}{3 - 1} = \dfrac{2(728)}{2} = 728$.
Q3 — Geometric Progression (G.P.) · medium · numerical
The sum to infinity of $8 + 4 + 2 + \cdots$ is:
A. $12$
B. $14$
C. $\infty$
D. $16$ ✓ Correct
Solution: $S_\infty = \dfrac{a}{1-r} = \dfrac{8}{1 - \tfrac12} = 16$.
Q4 — Geometric Progression (G.P.) · medium · numerical
Which term of the G.P. $2, 6, 18, \dots$ is $486$?
A. $5$-th
B. $4$-th
C. $7$-th
D. $6$-th ✓ Correct
Solution: $2 \cdot 3^{\,n-1} = 486 \Rightarrow 3^{\,n-1} = 243 = 3^5 \Rightarrow n = 6$.
Q5 — Geometric Progression (G.P.) · medium · numerical
If three geometric means are inserted between $2$ and $32$, the middle mean is:
A. $4$
B. $8$ ✓ Correct
C. $16$
D. $12$
Solution: With $5$ terms, $r^4 = \tfrac{32}{2} = 16 \Rightarrow r = 2$; the means are $4, 8, 16$, so the middle is $8$.
Q6 — Geometric Progression (G.P.) · medium · numerical
In a G.P. the $3$-rd term is $24$ and the $6$-th term is $192$. The common ratio is:
A. $4$
B. $3$
C. $\tfrac12$
D. $2$ ✓ Correct
Solution: $\dfrac{ar^5}{ar^2} = \dfrac{192}{24} = 8 = r^3 \Rightarrow r = 2$.
Q7 — Geometric Progression (G.P.) · medium · numerical
The sum to infinity of $1 - \tfrac13 + \tfrac19 - \cdots$ is:
A. $\dfrac43$
B. $\dfrac34$ ✓ Correct
C. $\dfrac23$
D. $\dfrac32$
Solution: $a = 1, r = -\tfrac13$: $S_\infty = \dfrac{1}{1 + \tfrac13} = \dfrac34$.
Q8 — Geometric Progression (G.P.) · hard · numerical
If the sum of the first $n$ terms of the G.P. $5 + 10 + 20 + \cdots$ is $635$, then $n$ equals:
A. $9$
B. $6$
C. $7$ ✓ Correct
D. $8$
Solution: $\dfrac{5(2^n - 1)}{2-1} = 635 \Rightarrow 2^n - 1 = 127 \Rightarrow 2^n = 128 \Rightarrow n = 7$.
Q9 — Geometric Progression (G.P.) · medium · numerical
Three numbers are in G.P. with product $512$. Their middle term is:
A. $16$
B. $8$ ✓ Correct
C. $64$
D. $4$
Solution: For $\tfrac{a}{r}, a, ar$ the product is $a^3 = 512 \Rightarrow a = 8$.
Q10 — Geometric Progression (G.P.) · medium · numerical
The number of terms in the G.P. $3, 6, 12, \dots, 384$ is:
A. $7$
B. $8$ ✓ Correct
C. $6$
D. $9$
Solution: $3 \cdot 2^{\,n-1} = 384 \Rightarrow 2^{\,n-1} = 128 \Rightarrow n = 8$.
Q11 — Geometric Progression (G.P.) · hard · numerical
The geometric mean of the five numbers $2, 4, 8, 16, 32$ is:
A. $10$
B. $12$
C. $16$
D. $8$ ✓ Correct
Solution: G.M. $= (2 \cdot 4 \cdot 8 \cdot 16 \cdot 32)^{1/5} = (2^{15})^{1/5} = 2^3 = 8$.
Q12 — Geometric Progression (G.P.) · hard · numerical
The sum to infinity of a G.P. is $15$ and the sum of the squares of its terms is $45$. Its first term is:
A. $5$ ✓ Correct
B. $3$
C. $9$
D. $6$
Solution: $\dfrac{a}{1-r} = 15$ and $\dfrac{a^2}{1-r^2} = 45$; dividing gives $\dfrac{a}{1+r} = 3$, so $r = \tfrac23, a = 5$.
Q13 — Geometric Progression (G.P.) · hard · numerical
The value of $9^{1/3} \cdot 9^{1/9} \cdot 9^{1/27} \cdots$ (to infinity) is:
A. $3$ ✓ Correct
B. $9$
C. $27$
D. $\sqrt{3}$
Solution: Exponent $= \tfrac13 + \tfrac19 + \cdots = \dfrac{1/3}{1 - 1/3} = \tfrac12$, so the product is $9^{1/2} = 3$.
Q14 — Geometric Progression (G.P.) · hard · numerical
In a G.P., the sum of the first two terms is $12$ and the sum of the next two terms is $48$. The common ratio (positive) is:
A. $\tfrac12$
B. $3$
C. $2$ ✓ Correct
D. $4$
Solution: $(ar^2 + ar^3) = r^2(a + ar) \Rightarrow 48 = 12r^2 \Rightarrow r^2 = 4 \Rightarrow r = 2$.
Q15 — Geometric Progression (G.P.) · medium · numerical
If the $2$-nd term of a G.P. is $3$ and the $5$-th term is $81$, the common ratio is:
A. $3$ ✓ Correct
B. $\tfrac13$
C. $27$
D. $2$
Solution: $\dfrac{ar^4}{ar} = \dfrac{81}{3} = 27 = r^3 \Rightarrow r = 3$.
Q16 — Geometric Progression (G.P.) · medium · theory
If $x, y, z$ are in G.P., then:
A. $y^2 = x + z$
B. $y^2 = xz$ ✓ Correct
C. $2y = x + z$
D. $y = x + z$
Solution: The middle term of a G.P. is the geometric mean of its neighbours: $y^2 = xz$.
Q17 — Geometric Progression (G.P.) · medium · theory
If $a, b, c$ are in G.P., then $\dfrac1a, \dfrac1b, \dfrac1c$ are in:
A. no progression
B. H.P.
C. A.P.
D. G.P. ✓ Correct
Solution: Taking reciprocals of a G.P. gives another G.P. (with ratio $1/r$).
Q18 — Geometric Progression (G.P.) · medium · numerical
The sum to infinity of $\tfrac12 + \tfrac14 + \tfrac18 + \cdots$ is:
A. $\tfrac34$
B. $1$ ✓ Correct
C. $\tfrac12$
D. $2$
Solution: $S_\infty = \dfrac{1/2}{1 - 1/2} = 1$.
Q19 — Geometric Progression (G.P.) · hard · numerical
For a G.P. with first term $a$ and ratio $r$ ($|r|<1$), if $S_\infty = 4$ and $a = 3$, then $r$ is:
A. $\tfrac13$
B. $\tfrac34$
C. $\tfrac14$ ✓ Correct
D. $\tfrac12$
Solution: $\dfrac{3}{1-r} = 4 \Rightarrow 1 - r = \tfrac34 \Rightarrow r = \tfrac14$.
Q20 — Geometric Progression (G.P.) · hard · numerical
If $a, b, c, d$ are in G.P., then $(b - c)^2 + (c - a)^2 + (d - b)^2$ equals:
A. $(b - c)^2$
B. $(a + d)^2$
C. $0$
D. $(a - d)^2$ ✓ Correct
Solution: A standard G.P. identity: $(b-c)^2 + (c-a)^2 + (d-b)^2 = (a-d)^2$.