Arithmetico-Geometric Progression (A.G.P.) — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Arithmetico-Geometric Progression (A.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 — Arithmetico-Geometric Progression (A.G.P.) · medium · theory
The $n$-th term of the A.G.P. $a + (a+d)r + (a+2d)r^2 + \cdots$ is:
A. $[a + (n-1)d] + r^{\,n-1}$
B. $a\,d^{\,n-1} r$
C. $[a + (n-1)d]\, r^{\,n-1}$ ✓ Correct
D. $(a + nd)\, r^n$
Solution: The $n$-th term multiplies the A.P. term $a + (n-1)d$ by the G.P. factor $r^{\,n-1}$.
Q2 — Arithmetico-Geometric Progression (A.G.P.) · medium · theory
The series $1 + 2x + 3x^2 + 4x^3 + \cdots$ is an A.G.P. with:
A. $a = 0, d = 1, r = x$
B. $a = 1, d = 2, r = x$
C. $a = 1, d = 1, r = 2$
D. $a = 1, d = 1, r = x$ ✓ Correct
Solution: The A.P. part is $1, 2, 3, \dots$ ($a = 1, d = 1$) and the G.P. part has ratio $x$.
Q3 — Arithmetico-Geometric Progression (A.G.P.) · hard · numerical
For $|x| < 1$, the sum to infinity of $1 + 2x + 3x^2 + 4x^3 + \cdots$ is:
A. $\dfrac{1}{(1+x)^2}$
B. $\dfrac{1}{1-x}$
C. $\dfrac{x}{(1-x)^2}$
D. $\dfrac{1}{(1-x)^2}$ ✓ Correct
Solution: This is the standard A.G.P./binomial sum $\sum_{n\ge 1} n x^{\,n-1} = \dfrac{1}{(1-x)^2}$.
Q4 — Arithmetico-Geometric Progression (A.G.P.) · hard · numerical
The sum to infinity of $1 + 2\left(\tfrac13\right) + 3\left(\tfrac13\right)^2 + \cdots$ is:
A. $\dfrac94$ ✓ Correct
B. $3$
C. $\dfrac32$
D. $\dfrac43$
Solution: Using $\dfrac{1}{(1-x)^2}$ with $x = \tfrac13$: $\dfrac{1}{(2/3)^2} = \dfrac94$.
Q5 — Arithmetico-Geometric Progression (A.G.P.) · hard · numerical
The sum $1 + 2 \cdot 2 + 3 \cdot 2^2 + 4 \cdot 2^3$ (first $4$ terms of $\sum k\, 2^{k-1}$) equals:
A. $49$ ✓ Correct
B. $33$
C. $48$
D. $50$
Solution: $1 + 4 + 12 + 32 = 49$ (matching the formula $(n-1)2^n + 1 = 3\cdot16+1$).
Q6 — Arithmetico-Geometric Progression (A.G.P.) · hard · numerical
The value of $\displaystyle\sum_{n=1}^{\infty} \dfrac{n}{2^n}$ is:
A. $4$
B. $2$ ✓ Correct
C. $\dfrac32$
D. $1$
Solution: Using $\sum n x^n = \dfrac{x}{(1-x)^2}$ with $x = \tfrac12$: $\dfrac{1/2}{1/4} = 2$.
Q7 — Arithmetico-Geometric Progression (A.G.P.) · hard · numerical
The value of $\displaystyle\sum_{n=1}^{\infty} \dfrac{n}{3^n}$ is:
A. $\dfrac34$ ✓ Correct
B. $\dfrac94$
C. $\dfrac13$
D. $\dfrac32$
Solution: $\sum n x^n = \dfrac{x}{(1-x)^2}$ with $x = \tfrac13$: $\dfrac{1/3}{4/9} = \dfrac34$.
Q8 — Arithmetico-Geometric Progression (A.G.P.) · medium · numerical
In the A.G.P. $2 \cdot 1 + 5 \cdot 2 + 8 \cdot 4 + \cdots$ (A.P. part $2, 5, 8$; G.P. ratio $2$), the $3$-rd term is:
A. $16$
B. $40$
C. $32$ ✓ Correct
D. $24$
Solution: Third term $= (8)(2^2) = 8 \cdot 4 = 32$.
Q9 — Arithmetico-Geometric Progression (A.G.P.) · medium · numerical
For the A.G.P. $1 \cdot 1 + 3 \cdot 2 + 5 \cdot 4 + 7 \cdot 8 + \cdots$, the $5$-th term is:
A. $128$
B. $72$
C. $144$ ✓ Correct
D. $160$
Solution: A.P. part $1,3,5,7,9$; G.P. part $1,2,4,8,16$; the $5$-th term is $9 \cdot 16 = 144$.
Q10 — Arithmetico-Geometric Progression (A.G.P.) · hard · numerical
The sum to infinity of $3 + 5\left(\tfrac12\right) + 7\left(\tfrac12\right)^2 + \cdots$ is:
A. $8$
B. $9$
C. $10$ ✓ Correct
D. $12$
Solution: With $a = 3, d = 2, r = \tfrac12$: $S_\infty = \dfrac{a}{1-r} + \dfrac{dr}{(1-r)^2} = 6 + 4 = 10$.
Q11 — Arithmetico-Geometric Progression (A.G.P.) · hard · numerical
The sum to infinity of $1 + \dfrac32 + \dfrac54 + \dfrac78 + \cdots$ is:
A. $4$
B. $8$
C. $6$ ✓ Correct
D. $5$
Solution: $a = 1, d = 2, r = \tfrac12$: $S_\infty = \dfrac{1}{1/2} + \dfrac{2 \cdot \tfrac12}{(1/2)^2} = 2 + 4 = 6$.
Q12 — Arithmetico-Geometric Progression (A.G.P.) · hard · numerical
The value of $1 + 4\left(\tfrac12\right) + 7\left(\tfrac12\right)^2 + 10\left(\tfrac12\right)^3 + \cdots$ (to infinity) is:
A. $4$
B. $10$
C. $6$
D. $8$ ✓ Correct
Solution: $a = 1, d = 3, r = \tfrac12$: $S_\infty = \dfrac{1}{1/2} + \dfrac{3 \cdot \tfrac12}{(1/2)^2} = 2 + 6 = 8$.
Q13 — Arithmetico-Geometric Progression (A.G.P.) · medium · numerical
In the A.G.P. $2 + 6 \cdot 3 + 10 \cdot 9 + \cdots$ (A.P. part $2, 6, 10$; G.P. ratio $3$), the $4$-th term is:
A. $378$ ✓ Correct
B. $126$
C. $252$
D. $405$
Solution: A.P. term $= 14$, G.P. factor $= 3^3 = 27$; product $= 14 \cdot 27 = 378$.
Q14 — Arithmetico-Geometric Progression (A.G.P.) · medium · numerical
The sum $1 + 2 \cdot 3 + 3 \cdot 3^2$ (first three terms of $\sum k\, 3^{k-1}$) equals:
A. $30$
B. $34$ ✓ Correct
C. $36$
D. $40$
Solution: $1 + 6 + 27 = 34$.
Q15 — Arithmetico-Geometric Progression (A.G.P.) · medium · numerical
The $n$-th term of the sequence $1, 2\cdot 2, 3\cdot 2^2, \dots$ is $n\, 2^{\,n-1}$; its $4$-th term is:
A. $48$
B. $16$
C. $24$
D. $32$ ✓ Correct
Solution: $T_4 = 4 \cdot 2^3 = 4 \cdot 8 = 32$.
Q16 — Arithmetico-Geometric Progression (A.G.P.) · hard · numerical
For $|x| < 1$, the sum $1 - 2x + 3x^2 - 4x^3 + \cdots$ equals:
A. $\dfrac{1}{1+x}$
B. $\dfrac{x}{(1+x)^2}$
C. $\dfrac{1}{(1-x)^2}$
D. $\dfrac{1}{(1+x)^2}$ ✓ Correct
Solution: Replacing $x$ by $-x$ in $\dfrac{1}{(1-x)^2} = \sum n x^{\,n-1}$ gives $\dfrac{1}{(1+x)^2}$.
Q17 — Arithmetico-Geometric Progression (A.G.P.) · hard · numerical
The value of $\displaystyle\sum_{n=1}^{\infty} (2n+1)\left(\tfrac13\right)^n$ is:
A. $\dfrac32$
B. $\dfrac52$
C. $3$
D. $2$ ✓ Correct
Solution: $2\sum n x^n + \sum x^n$ with $x = \tfrac13$: $2 \cdot \dfrac{1/3}{4/9} + \dfrac{1/3}{2/3} = \dfrac32 + \dfrac12 = 2$.
Q18 — Arithmetico-Geometric Progression (A.G.P.) · medium · theory
The standard technique to sum an A.G.P. of $n$ terms is to:
A. take reciprocals of all terms
B. add $r$ to each term
C. differentiate the series
D. multiply the sum $S$ by the ratio $r$ and subtract from $S$ ✓ Correct
Solution: Forming $S - rS$ collapses the A.G.P. into a G.P. plus a boundary term, which is then easily summed.
Q19 — Arithmetico-Geometric Progression (A.G.P.) · hard · numerical
The value of $\displaystyle\sum_{n=1}^{\infty} \dfrac{n^2}{2^n}$ is:
A. $8$
B. $4$
C. $6$ ✓ Correct
D. $2$
Solution: Using $\sum n^2 x^n = \dfrac{x(1+x)}{(1-x)^3}$ with $x = \tfrac12$: $\dfrac{(1/2)(3/2)}{(1/2)^3} = \dfrac{3/4}{1/8} = 6$.
Q20 — Arithmetico-Geometric Progression (A.G.P.) · hard · numerical
The sum of the first $n$ terms of the A.G.P. $1 + 2x + 3x^2 + \cdots + n x^{\,n-1}$ (for $x \ne 1$) is:
A. $\dfrac{1 - x^n}{(1-x)^2}$
B. $\dfrac{1 - x^{\,n+1}}{1-x}$
C. $\dfrac{n x^n}{1-x}$
D. $\dfrac{1 - (n+1)x^n + n x^{\,n+1}}{(1-x)^2}$ ✓ Correct
Solution: Using the $S - xS$ method, $S_n = \dfrac{1 - (n+1)x^n + n x^{\,n+1}}{(1-x)^2}$, which tends to $\dfrac{1}{(1-x)^2}$ as $n \to \infty$ for $|x|<1$.