Question

If the sum of the second, third and fourth terms of a positive term G.P. is $3$ and the sum of its sixth, seventh and eighth terms is $243$, then the sum of the first $50$ terms of this G.P. is:

• A. $\frac{2}{13}\left(3^{50}-1\right)$
• B. $\frac{1}{26}\left(3^{49}-1\right)$
• C. $\frac{1}{13}\left(3^{50}-1\right)$
• D. $\frac{1}{26}\left(3^{50}-1\right)$

Answer 1

The correct option is D

$\frac{1}{26}\left(3^{50}-1\right)$

Explanation for the correct option.

Step 1: Find the value of $r$.

Let the terms of G.P be $a,ar,a{r}^2,.........$

Then $ar+a{r}^2+a{r}^3=3........\left(1\right)$ and

$\begin{array}{rcl}a{r}^5+a{r}^6+a{r}^7& =& 243\\ & \Rightarrow & r^4\left(ar+a{r}^2+a{r}^3\right)=243\\ \Rightarrow 3r^4& =& 243\mathrm{from}\left(1\right)\\ \Rightarrow r^4& =& 3^4\\ \Rightarrow r& =& 3\end{array}$

Step 2: Find the value of $a$.

By substituting the value of $r$ in $\left(1\right)$, we get

$$\begin{gathered} a\left(3\right)+a{\left(3\right)}^2+a{\left(3\right)}^3=3 \\ \Rightarrow a\left(3+9+27\right)=3 \\ \Rightarrow a=\frac{1}{13} \end{gathered}$$

Step 3: Find the sum of first $50$ terms of given G.P.

The sum of $n$ terms of G.P $\mathbf{=}{\mathit{S}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{a}\left(r^n-1\right)}{\mathbf{r}\mathbf{-}\mathbf{1}}$.

$\begin{array}{rcl}& \Rightarrow & S_{50}=\frac{{\displaystyle \frac{1}{13}}(3^{50}-1)}{3-1}\\ & =& \frac{1}{26}\left(3^{50}-1\right)\end{array}$

Hence, option D is correct.