Question

In an A.P., prove that: `T_(m + n ) + T_(m - n) = 2.T_m`

Answer 1

Let a be the first term and d be the common difference. We define the three terms mentioned:

1. `T_(m + n): a + (m + n − 1)d`

2. `T_(m − n): a + (m − n − 1)d`

3. T_m: a + (m − 1)d

Add the terms on the Left Hand Side (LHS):

LHS = `T_(m + n) + T_(m − n)`

LHS = [a + (m + n − 1)d] + [a + (m − n − 1)d]

LHS = 2a + (m + n − 1 + m − n − 1)d

LHS = 2𝑎 + (2𝑚 − 2)𝑑

Factor out a 2 from the terms in the bracket

LHS = 2𝑎 + 2(𝑚 − 1)𝑑

LHS = 2[𝑎 + (𝑚 − 1)d]

a + (m − 1)d = T_m​

LHS = RHS

`T_(m + n ) + T_(m - n) = 2.T_m`

Hence Proved. This property shows that in an A.P., the sum of terms equidistant from T_m is twice the value of T_m.