Find k so that k-1, k, k+2 are consecutive terms of a G.P.

MISCELLANEOUS EXERCISE - 2 (II) · Q127

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Concept understanding — Geometric Progression

Geometric Progression: The Idea of Repeated Multiplication

Imagine you're folding a piece of paper in half. Start with thickness 1 unit. After one fold, thickness becomes 2. After two folds, thickness becomes 4. After three folds, thickness becomes 8. The sequence of thicknesses is:

1, 2, 4, 8, 16, ...

Notice the pattern: each term is obtained by multiplying the previous term by the same number (here, 2). That's the core intuition behind a geometric progression — you keep multiplying by a fixed number, step after step.

This is different from an arithmetic progression, where you keep adding a fixed number. Here, the growth is multiplicative, not additive. That's why geometric progressions grow (or shrink) much faster.

Precise Definition

A Geometric Progression (GP) is a sequence of numbers where the ratio of any term to its preceding term is constant. This constant is called the common ratio, denoted by $r$ .

If the first term is $a$ , then the sequence looks like:

a, \ ar, \ ar^2, \ ar^3, \ ar^4, \ \dots

Note

The common ratio $r$ can be any real number — positive, negative, or even a fraction. If $r$ is negative, the terms alternate in sign. If $0 < r < 1$ , the terms get smaller and smaller.

The $n$ -th Term

To find any term directly without listing all previous ones, use the formula:

T_n = a \cdot r^{\,n-1}

where $T_n$ is the $n$ -th term, $a$ is the first term, $r$ is the common ratio, and $n$ is the term number (starting from 1).

Example: For the paper-folding sequence, $a = 1$ , $r = 2$ . The 5th term is $1 \cdot 2^{5-1} = 2^4 = 16$ , which matches our list.

Sum of $n$ Terms

There are two cases, depending on whether $r = 1$ or not.

n

When $r = 1$ , every term is just $a$ , so the sum is simply $n \times a$ .

Why the formula works (intuition):

Let $S = a + ar + ar^2 + \dots + ar^{n-1}$ . Multiply both sides by $r$ : $rS = ar + ar^2 + \dots + ar^n$ . Subtract the first from the second: $rS - S = ar^n - a$ , so $S(r-1) = a(r^n - 1)$ , giving the formula above.

Sum of an Infinite GP

If the common ratio $r$ lies strictly between $-1$ and $1$ (i.e., $|r| < 1$ ), the terms get smaller and smaller, and the sum of all terms approaches a finite value:

S_\infty = \frac{a}{1 - r}, \quad \text{for } |r| < 1

Watch out

If $|r| \geq 1$ , the infinite sum does not exist (it diverges to infinity or oscillates without settling). Never apply the infinite sum formula when $|r| \geq 1$ .

Example: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ has $a=1$ , $r=\frac12$ , so $S_\infty = \frac{1}{1 - 1/2} = 2$ . This matches the intuition that repeatedly halving a unit length eventually fills exactly 2 units.

Quick Reference Table

Property	Formula	Condition
Common ratio	$r = \dfrac{T_{n+1}}{T_n}$	Always
$n$ -th term	$T_n = a r^{n-1}$	Always

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$k^2=(k-1)(k+2)=k^2+k-2 \Rightarrow 0=k-2 \Rightarrow …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 40 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{5}{2}, \dfrac{5}{4}, \dfrac{5}{8}, \ldots$
(a) $\dfrac{5}{2^{20}}$
(b) $\dfrac{1}{2^{20}}$
(c) $\dfrac{5}{2^{19}}$
(d) $\dfrac{5}{2^{18}}$
›Reveal solutionSolution

Identify the first term and common ratio, then apply $a_n = a\,r^{n-1}$ for $n=20$ .

The G.P. is $\dfrac52, \dfrac54, \dfrac58, \ldots$ with first term $a = \dfrac52$ and common ratio $r = \dfrac{5/4}{5/2} = \dfrac12$ .

The $n$ th term of a G.P. is $a_n = a\,r^{n-1}$ . For $n=20$ : …

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CBSE 2026Set ANNUAL1 markMCQ
$\sqrt{3}, 3, 3\sqrt{3}, \ldots$
(a) 9th
(b) 12th
(c) 13th
(d) 11th
›Reveal solutionSolution

The sequence is a G.P. with common ratio $\sqrt3$ ; express the $n$ th term as a power of $\sqrt3$ and match it to $729$ .

The sequence $\sqrt3, 3, 3\sqrt3, \ldots$ has first term $a=\sqrt3$ and common ratio $r = \dfrac{3}{\sqrt3} = \sqrt3$ (check: $3\sqrt3/3=\sqrt3$ too).

…

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CBSE 2026Set ANNUAL1 markMCQ
$a = 729$
(a) 2187
(b) 64
(c) 2159
(d) 2059
›Reveal solutionSolution

First find the common ratio from $a\,r^6 = 64$ , then sum the first 7 terms directly.

Given $a = 729$ and the 7th term $a_7 = a\,r^6 = 64$ :

$729\,r^6 = 64 \implies r^6 = \dfrac{64}{729} = \left(\dfrac23\right)^6 \implies r = \dfrac23$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The value of x for which the progression 2/7, x, 14 is in G.P., is:$
(a) 4
(b) 2
(c) 1
(d) None of these
›Reveal solutionSolution

In a G.P., each term squared (the middle term) equals the product of its neighbours; solving gives x = 2.

For $\dfrac{2}{7}, x, 14$ to be in G.P., the middle term squared must equal the product of the outer terms:

$x^2 = \frac{2}{7} \times 14 = 4$

$x = \pm 2$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The Geometric mean of 2 and 8 is:$
(a) $-4$
(b) $5$
(c) $-5$
(d) $4$
›Reveal solutionSolution

The geometric mean of two numbers $a,b$ is $\sqrt{ab}$ ; for 2 and 8 this gives $\sqrt{16}=4$ .

The geometric mean (GM) of two positive numbers $a$ and $b$ is defined as $G=\sqrt{ab}$ , the number such that $a, G, b$ form a G.P.

…

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CBSE 2026Set ANNUAL1 mark
$-\dfrac{2}{7}, x, -\dfrac{7}{2}$
›Reveal solutionSolution

If $a, x, c$ are consecutive terms of a G.P., then $x^2 = a\cdot c$ ; here that gives $x=\pm1$ , and $x=-1$ is the value consistent with a single common ratio across the progression.

For three consecutive G.P. terms $a, x, c$ : $\dfrac{x}{a} = \dfrac{c}{x} \Rightarrow x^2 = ac$ .

Here $a = -\dfrac{2}{7}$ , $c = -\dfrac{7}{2}$ , so $x^2 = \left(-\dfrac{2}{7}\right)\left(-\dfrac{7}{2}\right) = 1$ , giving $x = \pm 1$ .

Checking $x=-1$ : common ratio $r = \dfrac{x}{a} = \dfrac{-1}{-2/7} = \dfrac{7}{2}$ , and indeed $x \cdot r = -1 \times \dfrac{7}{2} = -\dfrac{7}{2} = c$ ✓ — a single positive ratio $\tfrac{7}{2}$ carries all three (negative) terms consistently.

…

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CBSE 2026Set 1A1 mark
$1, \frac{1}{3}, \frac{1}{9}, \ldots$
›Reveal solutionSolution

With $a=1$ , $r=\tfrac13$ , $S_\infty=\dfrac{1}{1-\frac13}=\dfrac32$ .

The GP $1,\tfrac13,\tfrac19,\dots$ has first term $a=1$ and common ratio $r=\tfrac13$ (with $|r|<1$ ). The sum to infinity is …

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CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{1}{9}, \dfrac{-1}{27}, \dfrac{1}{81}, \dfrac{-1}{243}, \ldots$
(a) $1/3$
(b) 3
(c) -3
(d) $-1/3$
›Reveal solutionSolution

The common ratio $r = -\dfrac{1}{3}$ .

For a G.P., $r = \dfrac{a_2}{a_1}$ .

Here $a_1 = \dfrac{1}{9}$ , $a_2 = -\dfrac{1}{27}$ : $r = \dfrac{-1/27}{1/9} = -\dfrac{1}{27}\times\dfrac{9}{1} = -\dfrac{9}{27} = -\dfrac{1}{3}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$1, 5, 25, 125, \ldots$
(a) $5^9$
(b) $5^{10}$
(c) $5^{11}$
(d) $5^{12}$
›Reveal solutionSolution

The 10th term is $5^9$ .

For the G.P. $1, 5, 25, 125, \ldots$ : first term $a=1$ , common ratio $r = 5/1 = 5$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The geometrical mean of 44 and 11 is$
(a) 27.5
(b) 25
(c) 55
(d) 22
›Reveal solutionSolution

The geometric mean of 44 and 11 is $22$ .

For two positive numbers $a$ and $b$ , the geometric mean is $\sqrt{ab}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$x$
(a) $x = 1, x = -1$
(b) $x = -1, x = 3$
(c) $x = 0, x = 2$
(d) None of these
›Reveal solutionSolution

For three numbers in G.P., the middle term squared equals the product of the outer terms: $x^2 = \left(-\frac27\right)\left(-\frac72\right) = 1$ , so $x=\pm1$ .

If $-\dfrac27, x, -\dfrac72$ are in G.P., then the middle term is the geometric mean of the other two: …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If third term of a G.P. is 24 and 6th term is 192, then its common ratio is:$
(a) 1/2
(b) 2
(c) 6
(d) None of these
›Reveal solutionSolution

Dividing the 6th term by the 3rd term of a G.P. gives $r^3$ .

Let the G.P. have first term $a$ and common ratio $r$ .

$a_3 = ar^2 = 24$ , $a_6 = ar^5 = 192$ .

…

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Q.Find kkk so that k−1,k,k+2k-1, k, k+2k−1,k,k+2 are consecutive terms of a G.P.

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Q.Find $k$ so that $k-1, k, k+2$ are consecutive terms of a G.P.

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