# 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

**Solution:**

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

It is known that the k^{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

Hence,

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 + 2 (m - 1) d

= 2 [a + (m - 1) d]

**= 2a _{m}**

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

NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 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

**Summary:**

We showed that the sum of (m + n)^{th} and (m - n)^{th} terms of an A.P is equal to twice the m^{th} term and we proved it using a_{k} = a + (k - 1) d