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# Show that the sum of $(m+n)^{th}$ and $(m-n)^{th}$ terms of an A.P is twice its $m^{th}$ term.

In an A.P. $n^{th}$ term = $t_n=a+(n-1)d$
$\Rightarrow\:t_{m+n}=a+(m+n-1)d$ ....(i) and
$t_{m-n}=a+(m-n-1)d.......$ (ii)
Adding (i) and (ii) we get
$t_{m+n}+t_{m-n}=(a+(m+n+1)d)+(a+(m-n-1)d)$
$\qquad\:\qquad=2a+(m+n-1+m-n-1)s$
$\qquad\qquad\:=2a+(2m-2)d=2(a+(m-1)d)=t_m$
Hence proved.