Question

The sum of $10$ terms of the series $12+16+24+40+.....$ is

• A. $4094$
• B. $4296$
• C. $4172$
• D. None of these

Answer 1

The correct option is C

$4172$

Explanation for correct option:

Let $S_n=12+16+24+40+...\mathrm{upto}10\mathrm{terms}$

$$\begin{gathered} S_n=\left(8+4\right)+(8+8)+(8+16)+(8+32)+.....\mathrm{upto}10\mathrm{terms} \\ {S}_n=\left[8\times 10+4+8+16+.....\mathrm{upto}10\mathrm{terms}\right] \end{gathered}$$

$4,8,16,......$ is a geometric progression with the first term $a=4$ and common ratio $r=2$.

Sum of geometric progression$=a\left[\frac{r^n-1}{r-1}\right]$

$4+8+16+......$$=4\left[\frac{{\left(2\right)}^{10}-1}{2-1}\right]$

$$\begin{gathered} =4\left[1024-1\right] \\ =4\times 1023 \\ =4092 \end{gathered}$$

$\begin{array}{rcl}& \Rightarrow & \left[80+4+8+16+......\right]=80+4092\\ & =& 4172\end{array}$

Hence, option C is correct.