Question

In an increasing geometric series, the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is $25$. Then, the sum of $4th,6thand8th$ terms is equal to:

• A. $35$
• B. $30$
• C. $26$
• D. $32$

Answer 1

The correct option is A

$35$

Explanation for the correct option.

Step 1: Solve for sum and product of the given terms.

The $nth$ term of GP is $t_n=a{r}^{n-1}$.

So, $t_2=ar\mathrm{and}{t}_6=a{r}^5$

$ar+a{r}^5=\frac{25}{2}...\left(1\right)$ and

$\begin{array}{rcl}a{r}^2\times a{r}^4& =& 25\\ \Rightarrow a^2{r}^6& =& 25\\ \Rightarrow a{r}^3& =& 5\\ \Rightarrow a& =& \frac{5}{r^3}\end{array}$

Substituting the value of $a$ in $\left(1\right)$, we get

$\begin{array}{rcl}\left(\frac{5}{r^3}\right)r+\left(\frac{5}{r^3}\right)r^5& =& \frac{25}{2}\\ \Rightarrow \frac{5}{r^2}+5r^2& =& \frac{25}{2}\\ \Rightarrow \frac{5\left(1+r^4\right)}{r^2}& =& \frac{25}{2}\\ \Rightarrow \frac{\left(1+r^4\right)}{r^2}& =& \frac{5}{2}\end{array}$

Step 2: Find the value of $r$.

$\begin{array}{rcl}2r^4-5r^2+2& =& 0\\ & \Rightarrow & 2r^4-4r^2-r^2+2=0\\ \Rightarrow \left(2r^2-1\right)\left(r^2-2\right)& =& 0\end{array}$

So, $r=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.\mathrm{or}\sqrt{2}$. As, it is increasing geometric series, so $r=\sqrt{2}$

Step 3: Find the sum of $4th,6thand8th$ terms.

$\begin{array}{rcl}a{r}^3+a{r}^5+a{r}^7& =& \frac{5}{2\sqrt{2}}\times 2\sqrt{2}+\frac{5}{2\sqrt{2}}\times 4\sqrt{2}+\frac{5}{2\sqrt{2}}\times 8\sqrt{2}\\ & =& 5+10+20\\ & =& 35\end{array}$

Hence, option A is correct.