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 $4^{\text{th}},6^{\text{th}}$ and $8^{\text{th}}$ terms is equal to :

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

Answer 1

The correct option is A $35$

Let $a$ be the first term and $r$ be the common ratio.
$ar+a{r}^5=\frac{25}{2}$
and $a{r}^2\times a{r}^4=25$
$\Rightarrow a^2{r}^6=25$
$\Rightarrow a{r}^3=5$
$\Rightarrow a=\frac{5}{r^3}\dots \left(1\right)$

Now, $\frac{5r}{r^3}+\frac{5r^5}{r^3}=\frac{25}{2}\left[\text{From (1)}\right]$
$\Rightarrow \frac{1}{r^2}+r^2=\frac{5}{2}$
Put $r^2=t$
$\Rightarrow \frac{t^2+1}{t}=\frac{5}{2}$
$\Rightarrow 2t^2-5t+2=0$
$\Rightarrow (2t-1)(t-2)=0$
$\Rightarrow t=\frac{1}{2},2$
$\Rightarrow r^2=\frac{1}{2},2$
$\Rightarrow r=\sqrt{2}$ as the G.P. is increasing.

$\therefore a{r}^3+a{r}^5+a{r}^7$
$=a{r}^3(1+r^2+r^4)$
$=5(1+2+4)=35$