Question

Find the ${10}^{th}$ term and sum of the 10 terms of the series $1,2,2^2,2^3,2^4....$

• A. 256, 512
• B. 512, 1023
• C. 256, 511
• D. 512, 1024

Answer 1

The correct option is B

512, 1023

Given series is a GP with common ratio = 2
$n^{th}$ term $t_n=a{r}^{n-1}$
${10}^{th}$ term $t_{10}={1.2}^{10-1}=2^9=512$
Sum of the n terms of GP

$s_n=\frac{a(r^n-1)}{r-1}$
Sum of the 10 terms of the series
Substitute n=10

$s_{10}=\frac{1.(2^{10}-1)}{2-1}=2^{10}-1=1023$