Question

The first and last terms of a G.P. are $a$ and $l$ respectively;$r$ being its common ratio; then the number of terms in this G.P is

• A. $\frac{(\log l–\log a)}{\log r}$
• B. $1-\frac{(\log l–\log a)}{\log r}$
• C. $\frac{(\log a–\log l)}{\log r}$
• D. $1+\frac{(\log l–\log a)}{\log r}$

Answer 1

The correct option is D

$1+\frac{(\log l–\log a)}{\log r}$

The explanation for the correct answer.

The G.P. is given as,

$a{r}^{n-1}=l$

$\frac{l}{a}=r^{n-1}$

Taking $\log$ both sides,

$\log \left(\frac{l}{a}\right)=(n-1)\log r\left[\mathbf{\because }\mathit{l}\mathit{o}\mathit{g}\left(\frac{m}{n}\right)\mathbf{=}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathit{m}\mathbf{}\mathbf{-}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathit{n}\right]$

$\log l–\log a+\log r=n\log r$

$\frac{\log l}{\log r}-\frac{\log a}{\log r}+\frac{\log r}{\log r}=n$

$\frac{\log l}{\log r}-\frac{\log a}{\log r}+1=n$

$\Rightarrow 1+\frac{(\log l–\log a)}{\log r}=n$

So, $n=1+\frac{(\log l–\log a)}{\log r}$

Hence, option (D) is the answer.