Question

The sum of the terms equidistant from the beginning and end in an A.P. is always same and is equal to the sum of ________ and ________ terms.

Answer 1

Let a₁, a₂, ............ a_n be an A.P with common difference d.
Then k^th term is a_k = a₁ + (k – 1)d and k^th term from last is the (n – k + 1)^th term from beginning
$$\begin{gathered} \mathrm{i}.\mathrm{e}{a}_{n-k+1}=a_1+\left(n-k+1-1\right)d \\ {a}_{n-k+1}=a_1+\left(n-k\right)d \end{gathered}$$
∴ k^th term from beginning + k^th term from the end is
$$\begin{gathered} a_1+\left(k-1\right)d+a_1+\left(n-k\right)d \\ =2a_1+d\left(k-1+n-k\right)d \\ =2a_1+\left(n-1\right)d \\ =a_1+\left(a_1+\left(n-1\right)d\right) \\ =a_1+a_n \end{gathered}$$
∴ Sum of the terms equidistant from the beginning of an A.P and end of A.P is always same and is equal to sum of first term and n^th term or last term of the A.P.