Question

The sums of first n terms of three A.P.s are S₁, S₂ and S₃. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that S₁ + S₃ = 2S₂.

Answer 1

Given: The sums of first n terms of three A.P.s are S₁, S₂ and S₃. The first term of each is 5 and their common differences are 2, 4 and 6 respectively.

To Prove: S₁ + S₃ = 2S₂

Proof [Step-wise]:

1. Formula: For an A.P. with first term a and common difference d, sum of first n terms is `S_n = n/2 [2a + (n - 1)d]`.

2. For the first A.P. (a = 5, d = 2):

`S_1 = n/2 [2 xx 5 + (n - 1) xx 2]` 

= `n/2 [10 + 2n - 2]`

= `n/2 [2n + 8]`

= n(n + 4)

3. For the second A.P. (a = 5, d = 4):

`S_2 = n/2 [2 xx 5 + (n - 1) xx 4]` 

= `n/2 [10 + 4n - 4]` 

= `n/2 [4n + 6]` 

= n(2n + 3)

4. For the third A.P. (a = 5, d = 6):

`S_3 = n/2 [2 xx 5 + (n - 1) xx 6]` 

= `n/2 [10 + 6n - 6]` 

= `n/2 [6n + 4]` 

= n(3n + 2)

5. Add S₁ and S₃: 

S₁ + S₃ = n(n + 4) + n(3n + 2)

= n[(n + 4) + (3n + 2)]

= n(4n + 6)

= 2n(2n + 3)

6. Compare with S₂: 

2S₂ = 2 × n(2n + 3) = 2n(2n + 3), which equals S₁ + S₃.

Thus S₁ + S₃ = 2S₂ as required.