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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 m th term.

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

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Let a be first term and d be common difference of AP.

Kth term of a AP is given by,

$a_k=a+(k-1)d$

$\therefore a_m_+_n=a+(m+n-1)d$

$\therefore a_m_-_n=a+(m-n-1)d$

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

$a_m_+_n+ a_m_-_n=a+(m+n-1)d+a+(m-n-1)d$

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

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

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

$=2.a_m$

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

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