Question

Show that the sum of (m + n)th and (m - n)th terms of an A.P. is equal to twice the mth terms.

Answer 1

Let $a_n$ be the first term and 'd' be the common difference of the given A.P.

$\therefore a_{m+n}=a+(m+n-1)d$ ..... (1)

$a_{m-n}=a+(m-n-1)d$ ..... (2)

Adding (1) and (2)

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

$=a+md+nd-d+a+md-nd-d$

$=2a+2md-2d$

$=2(a+md-d)$

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

Thus, sum of the (m + n)th and (m - n)th terms of an A.P. is equal to twice the mth term.