Question

The first term of a G. P. of $2n$ terms is $\text{'}a\text{'}$, and the last term is $\mathrm{I}$. The product of all the terms of the G. P. is

• A. ${\left(aI\right)}^{\frac{n}{2}}$
• B. ${\left(aI\right)}^{n-1}$
• C. ${\left(aI\right)}^n$
• D. ${\left(aI\right)}^{2n}$

Answer 1

The correct option is C

${\left(aI\right)}^n$

Step 1: Find the first term and the last term for the given G.P.

We have been given the first term of a G. P. of $2n$ terms is $\text{'}a\text{'}$ , and the last term is $\mathrm{I}$.

We need to find the product of all the terms of G.P.

We know the general form of a G.P. is $a,ar,a{r}^2,a{r}^3,a{r}^4,...,a{r}^{2n-1}$

where $a$represents the first term of the G.P.

and $r$ is the common ration

For given example,

the first term $=a$

Last term $$\begin{gathered} =a{r}^{2n-1} \\ =I \end{gathered}$$

and the number of terms $=2n$

Step 2: Find the required product.

The product of all terms of G.P would be,

$$\begin{gathered} \Rightarrow P=a\times ar\times a{r}^2\times ...\times a{r}^{2n-1} \\ \Rightarrow P=a^{2n}{r}^{(1+2+3+...+(2n-1\left)\right)}..............\left(i\right) \end{gathered}$$

The sum of the first $2n-1$terms would be,

$$\begin{gathered} \Rightarrow S_{2n-1}=\frac{(2n-1)(2n-1+1)}{2} \\ \Rightarrow S_{2n-1}=\frac{(2n-1)2n}{2} \\ \Rightarrow S_{2n-1}=(2n-1)n \end{gathered}$$

so, the equation $\left(i\right)$ would be,

$$\begin{gathered} \Rightarrow P=a^{2n}{r}^{(2n-1)n} \\ \Rightarrow P={(a\times a{r}^{2n-1})}^n \\ \Rightarrow P={(a\times I)}^n \\ \Rightarrow \mathit{P}\mathbf{=}{\mathbf{\left(}\mathit{a}\mathit{I}\mathbf{\right)}}^{\mathbf{n}} \end{gathered}$$

Therefore, option (C) is the correct answer.