Question

Sum of $(m+n)th$ and $(m-n)th$ term of an A.P. is equal to twice the mth term.

• A. True
• B. False

Answer 1

The correct option is A True

Let a be the first term and d be the common difference of the A.P. Then,

$a_{m+n}=(m+n)thterm=a+(m+n-1)d$

$a_{m-n}=(m-n)thterm=a+(m-n-1)d$

Therefore,

$a_{m+n}+a_{m-n}=[a+(m+n-1\left)d\right]+[a+(m-n-1\left)d\right]$

$=2a+(m+n-1+m-n-1)d$

$=2a+2(m-1)d$

$=2[a+(m-1\left)d\right]$

$=2a_m$