Question

If $a$ is the first term ,$1$ is the common difference and $b$ is the last term of an AP, then its sum is.

• A. $\frac{(a+b)(1+a-b)}{2}$
• B. $\frac{(a+b)(1-a+b)}{2}$
• C. $\frac{(a+b)(1-a)}{2}$
• D. $(a+b)(1-a+b)$

Answer 1

The correct option is B $\frac{(a+b)(1-a+b)}{2}$

Given first term of the A.P is ${}^{\prime }{a}^{\prime }$ common difference $1$ and last term is ${}^{\prime }{b}^{\prime }$
Thus number of terms $n=1+\frac{b-a}{1}=1+b-a$
Therefore, the sum is $=\frac{n}{2}(a+b)=\frac{(a+b)(1-a+b)}{2}$
Hence, option 'B' is correct.