Question

If $a_m$ denotes the $mth$ term of an A.P then $a_m$

• A. $\frac{2}{(a_{m+k}+a_{m-k})}$
• B. $\frac{(a_{m+k}–a_{m-k})}{2}$
• C. $\frac{(a_{m+k}+a_{m-k})}{2}$
• D. None of these

Answer 1

The correct option is C

$\frac{(a_{m+k}+a_{m-k})}{2}$

Explanation for the correct option:

Step 1. Find the value of $a_m$:

Let $a$ be the first term and $d$ be the common difference of AP.

$\therefore a_m=a+(m-1)d$

$a_{m+k}=a+(m+k-1)d$ …(i)

$a_{m-k}=a+(m-k-1)d$ ...(ii)

Step 2. By Adding equation (i) and (ii), we get

$\begin{array}{rcl}(a_{m+k}+a_{m-k})& =& 2a+d(m+k-1+m-k-1)\\ & =& 2a+d(m+k-1+m-k-1)\\ & =& 2a+d(2m-2)\\ & =& 2a+2d(m-1)\\ & =& 2\left[a+d(m-1)\right]\\ & =& 2a_m\end{array}$

$\mathbf{\therefore }{\mathit{a}}_{\mathbf{m}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{(}{\mathbf{a}}_{\mathbf{m}\mathbf{+}\mathbf{k}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{a}}_{\mathbf{m}\mathbf{-}\mathbf{k}}\mathbf{)}}{\mathbf{2}}$

Hence, Option ‘C’ is Correct.