Question

The sum of $10$ terms of GP $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots$ is

• A. $\frac{2^{10}-1}{2^{10}}$
• B. $\frac{2^9-1}{2^9}$
• C. $\frac{2^{10}-1}{2^9}$
• D. $\frac{2^9-1}{2^{10}}$

Answer 1

Answer: (A)

$\frac{2^{10}-1}{2^{10}}$

$S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ is in G.P

where $a=\frac{1}{2}$ and $r=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$

$S_n=\frac{a(r^n-1)}{r-1}$

$S_{10}=\frac{\frac{1}{2}({\left(\frac{1}{2}\right)}^{10}-1)}{\frac{1}{2}-1}$

$S_{10}=\frac{\frac{1}{2}({\left(\frac{1}{2}\right)}^{10}-1)}{\frac{-1}{2}}$

$=1-\frac{1}{2^{10}}$

$=\frac{2^{10}-1}{2^{10}}$