Question

A sequence $\left\{a_n\right\}$ is defined as
$a_n=\{\begin{array}{c}\frac{1}{n},\text{when}n\text{is odd}\\ -\frac{1}{n},\text{when}n\text{is even}\end{array}$
where $n\in N$. Then the sum of first four terms is

• A. $\frac{7}{12}$
• B. $\frac{5}{12}$
• C. $0$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $\frac{7}{12}$

$a_n=\{\begin{array}{c}\frac{1}{n},\text{when}n\text{is odd}\\ -\frac{1}{n},\text{when}n\text{is even}\end{array}$
$\therefore a_1=1,a_2=-\frac{1}{2},a_3=\frac{1}{3},a_4=-\frac{1}{4}$
Required sum $=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}=\frac{7}{12}$