Let $S$ be the sum, $P$ the product and $R$ the sum of reciprocals of $n$ terms in a G.P. Prove that $P^2 R^n = S^n$.

Question

Accepted Answer

> [!TLDR]
> For a geometric progression, the product $P$ is the $n$th power of the geometric mean, the sum $S$ is a standard finite sum, and the sum of reciprocals $R$ is a scaled version of $S$. Substituting these into $P^2 R^n$ simplifies directly to $S^n$, proving the identity.

The problem asks us to prove a relationship between three quantities defined from $n$ terms of a geometric progression (GP): the sum $S$, the product $P$, and the sum of reciprocals $R$. The identity $P^2 R^n = S^n$ looks symmetrical, and the key is to express each quantity in terms of the first term $a$ and the common ratio $r$.

Why does this work? Because a GP has a multiplicative structure — the product of terms is a simple power of $a$ and $r$, and the sum of reciprocals is just another GP with ratio $1/r$. The exponents then align beautifully.

Let the $n$ terms of the GP be:
$$a, ar, ar^2, \dots, ar^{n-1}$$

We will compute $S$, $P$, and $R$ one by one.

1. **Sum $S$ of the $n$ terms**  
   This is a standard finite geometric series:
   $$S = a + ar + ar^2 + \cdots + ar^{n-1} = a \cdot \frac{r^n - 1}{r - 1} \quad 	ext{(for } r 
eq 1	ext{)}$$
   If $r = 1$, all terms are $a$, so $S = na$, but the identity still holds (we'll check later). We'll assume $r 
eq 1$ for the main derivation.

2. **Product $P$ of the $n$ terms**  
   Multiply all terms:
   $$P = a \cdot ar \cdot ar^2 \cdots ar^{n-1} = a^n \cdot r^{0+1+2+\cdots+(n-1)}$$
   The exponent of $r$ is the sum of the first $n-1$ integers: $\frac{n(n-1)}{2}$. So:
   $$P = a^n r^{\frac{n(n-1)}{2}}$$

3. **Sum of reciprocals $R$**  
   The reciprocals of the terms are:
   $$\frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \dots, \frac{1}{ar^{n-1}}$$
   This is itself a GP with first term $\frac{1}{a}$ and common ratio $\frac{1}{r}$. Its sum is:
   $$R = \frac{1}{a} \cdot \frac{1 - (1/r)^n}{1 - 1/r} = \frac{1}{a} \cdot \frac{1 - r^{-n}}{1 - r^{-1}}$$
   Simplify the denominator: $1 - r^{-1} = \frac{r-1}{r}$. So:
   $$R = \frac{1}{a} \cdo

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

Q.Let $S$ be the sum, $P$ the product and $R$ the sum of reciprocals of $n$ terms in a G.P. Prove that $P^2 R^n = S^n$ .

Concept understanding — Geometric Progression

Geometric Progression: The Idea of Repeated Multiplication

Precise Definition

The $n$ -th Term

Sum of $n$ Terms

Sum of an Infinite GP

Quick Reference Table

Q.Let SSS be the sum, PPP the product and RRR the sum of reciprocals of nnn terms in a G.P. Prove that P2Rn=SnP^2 R^n = S^nP2Rn=Sn.

🎓Unlock everything free for 14 days

Q.Let $S$ be the sum, $P$ the product and $R$ the sum of reciprocals of $n$ terms in a G.P. Prove that $P^2 R^n = S^n$ .

Unlock everything free for 14 days