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

Question:

In a finite GP, prove that the product of the terms equidistant from the beginning and end is the product of first and last terms.

Solution:

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.