Question

# Find the geometric series whose ${5}^{\mathrm{th}}$ and ${8}^{\mathrm{th}}$ terms are $80$ and $640$, respectively.

Open in App
Solution

## Step 1. Find the common ratio:Here it is given that ${5}^{\mathrm{th}}$ term is $80$ and ${8}^{\mathrm{th}}$ term is $640$Let the first term is $a$ and the common ratio is$\mathrm{r}$ ${\mathrm{a}}_{5}={\mathrm{ar}}^{4}=80....\left(\mathrm{i}\right)$${\mathrm{a}}_{8}={\mathrm{ar}}^{7}=640....\left(\mathrm{ii}\right)$Divide (ii) and (i)$\begin{array}{rcl}& & \begin{array}{cc}â‡’& \frac{{\mathrm{ar}}^{7}}{{\mathrm{ar}}^{4}}=\frac{640}{80}\\ â‡’& {r}^{3}=8\\ â‡’& r=2\end{array}\end{array}$Step 2. Find the GP series:Put the value of $2$ in eq $\left(i\right)$$\begin{array}{rcl}& & \begin{array}{cc}â‡’& {\mathrm{ar}}^{4}=80\\ â‡’& aÃ—{2}^{4}=80\\ â‡’& a=\frac{80}{16}\\ â‡’& a=5\end{array}\end{array}$Second term,$â‡’\mathrm{ar}=5Ã—2=10$Third term $â‡’{\mathrm{ar}}^{2}=5Ã—{2}^{2}=20$Fourth term $â‡’{\mathrm{ar}}^{3}=5Ã—{2}^{3}=40$Therefore,the required GP is 5,10,20,40,80,..640.

Suggest Corrections
4
Join BYJU'S Learning Program
Select...
Related Videos
Geometric Progression
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
Select...