Question

In an $A.P.$ the sum of $m$ terms is equal ton and the sum of $n$ terms is equal to $m,$ then the sum of $(m+n)$ terms is:

• A. $m+n$
• B. $-(m+n)$
• C. $(m-n)$
• D. $-(m-n)$

Answer 1

The correct option is B $-(m+n)$

$\begin{array}{c}\text{Let first term of}A\cdot P\text{. be a and common}\\ \text{difference de}d\text{.}\\ \frac{m}{2}(2a+(m-1\left)d\right)=n\Rightarrow \text{lam}+\text{n}(m-1)d=2n\\ \text{and}\frac{n}{2}(2a+(n-1\left)d\right)=n\Rightarrow 2an+n(n-1)d=2m\end{array}$

$\begin{array}{cc}& \begin{array}{c}\text{Sustract both eqs.}\\ 2a(m-n)+[(m^2-n^2)-(m-n\left)\right]d=2(n-m)\end{array}\\ & \Rightarrow 2a(m-n)+(m-n)[m+n-1]d=-2(m-n)\\ & \Rightarrow 2a+(m+n-1)d=\frac{-2(m-n)}{(m-n)}=-2\\ & \left(\frac{w+x}{2}\right)(2a+(m+n-1\left)d\right)=-2\left(\frac{m+n}{2}\right)\\ & \text{option}B\text{is correct}\\ & =-(m+n)\end{array}$