Question

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

Answer 1

Step-by-step explanation:

To Prove :-

The sum of ( m+n)th and (m-n)th terms of an A.P. is equal to twice the mth term.

Proof:-

Let 'a' be the first term of term of the AP.

And 'd' be the common difference.

The nth term of an AP is:-

$\boxed{\sf \longrightarrow a_{n} = a + (n - 1)d}$

$\sf {(m + n)}^{th} \: term = a_{ m+ n} = a + (m + n - 1)d$

$\sf {(m - n)}^{th} \: term = a_{ m - n} = a + (m - n - 1)d$

Sum of (m + n)th term and (m - n)th term:-

$\sf \longrightarrow a_{ m + n} + a_{ m - n}$

$\sf \longrightarrow \big[a + (m + n - 1)d \big]+ \big[a + (m - n - 1)d \big]$

$\sf \longrightarrow a + a + (m + n - 1)d + (m - n - 1)d$

$\sf \longrightarrow 2a + (m + n - 1 + m - n - 1)d$

$\sf \longrightarrow 2a + (2m - 2)d$

$\sf \longrightarrow 2(a + (m - 1)d$

$\sf \longrightarrow2( a_{m})$

$\sf \longrightarrow2( {m}^{th} \: term)$

Hence Proved

Answer 2

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