Question

Sum of series $\left(1+2\right)+\left(3+5\right)+\left(6+7\right)+\left(9+10\right)+...+\left(93+94\right)+\left(95+97\right)+\left(98+99\right)$ will be:

• A. $5050$
• B. $3750$
• C. $1350$
• D. $4250$

Answer 1

The correct option is B

$3750$

The explanation for the correct option (Option B):

Sum of first $n$ integers is given by $\frac{n\left(n+1\right)}{2}$.

Let $S$ be the value of $\left(1+2\right)+\left(3+5\right)+\left(6+7\right)+\left(9+10\right)+...+\left(93+94\right)+\left(95+97\right)+\left(98+99\right)$.

$$\begin{gathered} \Rightarrow S=\left(1+2+3+4+...+100\right)-\left(4+8+12+16+...+100\right)[Rearrangingthetermsintotwoseries] \\ \Rightarrow S=\left(1+2+3+4+...+100\right)-4\left(1+2+3+4+...+25\right) \\ \Rightarrow S=\frac{100\times \left(100+1\right)}{2}-4\times \frac{25\times \left(25+1\right)}{2} \\ \Rightarrow S=50\times 101-50\times 26 \\ \Rightarrow S=50\times \left(101-26\right) \\ \Rightarrow S=50\times 75 \\ \Rightarrow S=3750 \end{gathered}$$

Hence, the correct answer is option B.