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

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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