Question

Sum $-\frac{1}{3},\frac{1}{2},-\frac{3}{4}$, .... to $7$ terms.

Answer 1

In the given series, $-\frac{1}{3},\frac{1}{2},-\frac{3}{4},...$

The ratio of first two terms is $\frac{\frac{1}{2}}{-\frac{1}{3}}=-\frac{3}{2}$

The ratio of third term to the second term is $\frac{-\frac{3}{4}}{\frac{1}{2}}=-\frac{3}{2}$

Hence, the terms are in Geometric Progression, with first term $a=-\frac{1}{3}$ and common ratio $r=-\frac{3}{2}$

Sum of terms in a GP is $\frac{a(1-r^n)}{(1-r)}=\frac{-\frac{1}{3}(1-{\left(\frac{-3}{2}\right)}^n)}{1-(-\frac{3}{2})}$

Here, $n=7\Rightarrow \text{Sum}=\frac{-\frac{1}{3}(1+{\left(\frac{3}{2}\right)}^7)}{1+\frac{3}{2}}=-\frac{1}{3}\times \frac{2}{5}\left(\frac{3^7+2^7}{2^7}\right)=-\frac{463}{192}$