Question

If $\frac{c}{2}+\frac{2c}{2^2}+\frac{3c}{2^3}=\frac{1}{2}$, then find $c$.

• A. $\frac{24}{11}$
• B. $\frac{4}{11}$
• C. $\frac{14}{11}$
• D. $-\frac{4}{11}$

Answer 1

Answer: (B)

$\frac{4}{11}$

$\frac{c}{2}+\frac{2c}{2^2}+\frac{3c}{2^3}=\frac{1}{2}$

$\frac{2^2\times c}{2^3}+\frac{2\times 2c}{2^3}+\frac{3c}{2^3}=\frac{1}{2}$

Making same denominator.

$\frac{4c+4c+3c}{2^3}=\frac{1}{2}$

$11c=\frac{2^3}{2}$

$c=\frac{2^2}{11}=\frac{4}{11}$.