Question

If the second and fifth terms of a GP are $24$ and $3$ respectively, then the sum of the first six terms is

• A. $181$
• B. $\frac{181}{2}$
• C. $189$
• D. $\frac{189}{2}$

Answer 1

The correct option is D

$\frac{189}{2}$

Step 1: Solve for first term and common ratio

Given that second and fifth terms of a GP are $24$ and $3$ respectively i.e. $t_2=24$ and $t_5=3$

We know that $t_n=a.r^{n-1}$ where, $a$ is the first term and $r$ is the common ratio

$$\begin{gathered} \therefore t_2=a.r^{2-1} \\ =ar \\ \Rightarrow ar=24...\left[1\right] \end{gathered}$$

$$\begin{gathered} t_5=a.r^{5-1} \\ =a{r}^4 \\ \Rightarrow a{r}^4=3...\left[2\right] \end{gathered}$$

On dividing $\left[2\right]$ by $\left[1\right]$, we get,

$\begin{array}{crcl}& \frac{a{r}^4}{ar}& =& \frac{3}{24}\\ \Rightarrow & r^3& =& \frac{1}{8}\\ \Rightarrow & r& =& \frac{1}{2}\\ \therefore & a\left(\frac{1}{2}\right)& =& 24\\ \Rightarrow & a& =& 48\end{array}$

Step 2: Solve for sum of first six terms

We know that sum of $n$terms of a G.P. is $S_n=\frac{a\left(1-r^n\right)}{1-r}$

$\begin{array}{cccl}\therefore & S_6& =& \frac{48\left(1-{\left({\displaystyle \frac{1}{2}}\right)}^6\right)}{1-{\displaystyle \frac{1}{2}}}\\ & & =& \frac{48\left({\displaystyle \frac{64-1}{64}}\right)}{{\displaystyle \frac{2-1}{2}}}\\ & & =& \frac{48\left({\displaystyle \frac{63}{64}}\right)}{{\displaystyle \frac{1}{2}}}\\ & & =& \frac{48\times 63\times 2}{64}\\ \Rightarrow & S_6& =& \frac{189}{2}\end{array}$

Hence, the correct option is option (D) i.e. $\frac{189}{2}$.