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

Question:

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

Solution:

Let $a$ and $d$ be the first term and the common difference of the A.P. respectively.

It is known that the $k^{\text {th }}$ 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 $(m+n)^{\text {th }}$ and $(m-n)^{\text {th }}$ terms of an A.P. is equal to twice the $m^{\text {th }}$ term.