Question

An A.P. consists of n (odd terms) and its middle term is m. Then the sum of the A.P. is

• A. 2 mn
• B. 1/2 mn
• C. mn
• D. $mn^2$.

Answer 1

Answer: (C)

mn

Answer 2

Middle term,
$\frac{T_{n+1}}{2} =m$
$⇒a+\left(\frac{n+1}{2} -1\right)d = m$
$\Rightarrow 2a+\left(n+1-2\right)d = 2m$
$\Rightarrow 2a+\left(n-1\right)d = 2m$
$\Rightarrow \frac{n}{2}\left[2a+\left(n-1\right)d\right]= \frac{n}{2}\left(2m\right) = mn$
$\Rightarrow \frac{n}{2}\left[2a+\left(n-1\right)d\right]= mn$
$\Rightarrow S_{n}$ of $A.P. = mn$

So, the correct option is (C): $mn$.