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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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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