Question

prove that in any ap am+n + am-n = 2am

Answer 1

Hope this helps you.....

Answer 2

Answer:

$a_{m+n}+a_{m-n} = 2a_{m}$

Step-by-step explanation:

Let a and d are first term and

common difference of an A.P.

$We \: know \:that ,\\\boxed {n^{th}\: term =a_{n}=a+(n-1)d}$

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

$ii) a_{m-n} = a+(m-n-1)d \: ---(2)$

/* Add equation (1) and (2) , we get

$LHS = a_{m+n}+a_{m-n}\\=a+(m+n-1)d+a+(m-n-1)\\=2a+(m+n-1+m-n-1)d\\=2a+(2m-2)d\\=2a+2(m-1)d\\=2[a+(m-1)d]\\=2a_{m}\\=RHS$

Therefore,

$a_{m+n}+a_{m-n} = 2a_{m}$

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