Show that the sum of $(m+n)^{ th }$ and $(m-n)^{ th }$ terms of an $A.P.$ is equal to twice the $m^{\text {th }}$ term.
Show that the sum of $(m+n)^{ th }$ and $(m-n)^{ th }$ terms of an $A.P.$ is equal to twice the $m^{\text {th }}$ term.
Let $a$ and $d$ be the first term and the common difference of the $A.P.$ respectively. It is known that the $k^{th}$ term of an $A.P.$ is given by
$a_{k}=a+(k-1) d$
$\therefore a_{m+n}=a+(m+n-1) d$
$a_{m-n}=a+(m-n-1) d$
$a_{m}=a+(m-1) d$
$\therefore a_{m+n}+a_{m-n}=a+(m+n-1) d+a+(m-n-1) d$
$=2 a+(m+n-1+m-n-1) d$
$=2 a+(2 m-2) d$
$=2 a+2(m-1) d$
$=2[a+(m-1) d]$
$=2 a_{m}$
Thus, the sum of $(m+n)^{t h}$ and $(m-n)^{t h}$ terms of an $A.P.$ is equal to twice the $m^{\text {th }}$ term.
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