Question

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

Answer 1

first term= a
common difference= d
than 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
=2a +(m-1) 2d
= 2am

Answer 2

Answer and Explanation:

To show : The sum of (m+n)th and (m-n)th term of an A.P is equal to twice the m th term?

Solution :

Let the A.P be $a,a+d,2a+d,..$

The first term is 'a'

The common difference is 'd'.

The nth term of the A.P is given by, $a_n=a+(n-1)d$

The (m+n)th term of A.P is $a_{m+n}=a+((m+n)-1)d=a+(m+n-1)d$

The (m-n)th term of A.P is $a_{m-n}=a+((m-n)-1)d=a+(m-n-1)d$

The sum of  (m+n)th and (m-n)th term of an A.P is given by,

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

$a_{m+n}+a_{m-n}=a+md+nd-d+a+md-nd-d$

$a_{m+n}+a_{m-n}=2a+2md-2d$

$a_{m+n}+a_{m-n}=2(a+md-d)$

$a_{m+n}+a_{m-n}=2(a+(m-1)d)$ ....(1)

Now, m th term of A.P is given by,

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

Twice of m th term is $2a_m=2(a+(m-1)d)$

From (1),

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

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