If the pth and qth terms of a G.P. are q and p respectively, show that its (p+q)th term is ((qp)/(pq))(1)/(p-q

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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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When the $p$ th term equals $q$ and the $q$ th term equals $p$ , we use the G.P. formula to set up two equations in $a$ and $r$ , then eliminate $a$ to find $r^{p-q}$ . Substituting back gives the $(p+q)$ th term as $\left(\dfrac{q^p}{p^q}\right)^{\frac{1}{p-q}}$ .

The heart of this problem lies in recognizing that a geometric progression is completely determined by its first term and common ratio. When we're told specific terms take specific values, we're really being given a system of equations that pins down these two parameters. The twist here is that the positions and values are swapped in a symmetric way, which creates an elegant relationship.

For any G.P. with first term $a$ and common ratio $r$ , the $n$ th term is $a \cdot r^{n-1}$ . This formula encodes how each term is built by multiplying the first term by the ratio raised to one less than the position number.

Let's translate what we know into equations and work systematically toward the $(p+q)$ th term.

1. Write the given conditions as equations

The $p$ th term is $q$ :

a \cdot r^{p-1} = q \quad \text{...(i)}

The $q$ th term is $p$ :

a \cdot r^{q-1} = p \quad \text{...(ii)}

2. Eliminate $a$ to find a relationship for $r$

Divide equation (i) by equation (ii):

\frac{a \cdot r^{p-1}}{a \cdot r^{q-1}} = \frac{q}{p}

r^{(p-1)-(q-1)} = \frac{q}{p}

r^{p-q} = \frac{q}{p} \quad \text{...(iii)}

This tells us the common ratio raised to the power $(p-q)$ equals the ratio $\frac{q}{p}$ .

3. Find $a$ in terms of known quantities

From equation (i):

a = \frac{q}{r^{p-1}} \quad \text{...(iv)}

4. Express the $(p+q)$ th term

The $(p+q)$ th term is:

T_{p+q} = a \cdot r^{(p+q)-1} = a \cdot r^{p+q-1}

Substitute $a$ from equation (iv):

T_{p+q} = \frac{q}{r^{p-1}} \cdot r^{p+q-1} = q \cdot r^{(p+q-1)-(p-1)} = q \cdot r^{q}

5. Express $r^q$ using equation (iii)

From $r^{p-q} = \frac{q}{p}$ , we can write: …

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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.If the pppth and qqqth terms of a G.P. are qqq and ppp respectively, show that its (p+q)(p+q)(p+q)th term is (qppq)1p−q\left(\dfrac{q^p}{p^q}\right)^{\frac{1}{p-q}}(pqqp​)p−q1​.

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Q.If the $p$ th and $q$ th terms of a G.P. are $q$ and $p$ respectively, show that its $(p+q)$ th term is $\left(\dfrac{q^p}{p^q}\right)^{\frac{1}{p-q}}$ .

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