In a finite GP, prove that the product of the terms equidistant from the

We need to prove that the product of the terms equidistant from the beginning and end is the product of first and last terms in a finite GP.

Let us first consider a finite GP.

$\mathrm{A}, \mathrm{AR}, \mathrm{AR}^{2} \ldots \mathrm{AR}^{\mathrm{n}-1}, \mathrm{AR}^{\mathrm{n}}$

Where n is finite.

Product of first and last terms in the given GP $=A \cdot A R^{n}$

$=\mathrm{A}^{2} \mathrm{R}^{n} \rightarrow(\mathrm{a})$

Now, $\mathrm{n}^{\text {th }}$ term of the GP from the beginning $=\mathrm{AR}^{\mathrm{n}-1} \rightarrow$ (1)

Now, starting from the end,

First term $=\mathrm{AR}^{n}$

Last term = A

$\frac{1}{R}=$ Common Ratio

So, an $\mathrm{n}^{\text {th }}$ term from the end of $\mathrm{GP}, \mathrm{A}_{\mathrm{n}}=\left(\mathrm{AR}^{\mathrm{n}}\right)\left(\frac{1}{\mathrm{R}^{\mathrm{n}-1}}\right)=\mathrm{AR} \rightarrow$ (2)

So, the product of $n^{\text {th }}$ terms from the beginning and end of the considered GP from (1) and $(2)=\left(\mathrm{AR}^{\mathrm{n}-1}\right)(\mathrm{AR})$

$=\mathrm{A}^{2} \mathrm{R}^{n} \rightarrow(\mathrm{b})$

So, from (a) and (b) its proved that the product of the terms equidistant from the beginning and end is the product of first and last terms in a finite GP.