If a, b, c, d are in A.P.; a, b, c are in G.P.; and 1/a, 1/b, 1/c are in A.P., prove that a, b, c are in G.P.

The given series is 3, 7, 13, 21, 31, …

On subtracting both equations, we obtain:

S − S = [3 + 7 + 13 + 21 + 31 + … + aₙ] − [3 + 7 + 13 + 21 + 31 + … + aₙ₋₁]

⇒ S − S = aₙ − aₙ₋₁

Let aₙ = 3 + [4 + 6 + 8 + … + (n − 1) terms]

⇒ aₙ = 3 + 2n(n − 1)

⇒ aₙ = n² + n + 1

Now, sum of n terms:

Sₙ = Σaₙ = Σ(n² + n + 1)

= Σn² + Σn + Σ1

= [n(n + 1)(2n + 1)/6] + [n(n + 1)/2] + n

= [n/3](n² + 3n + 5)

If S₁, S₂, S₃ are the sums of first n natural numbers, their squares, and their cubes respectively, then:

S₃ = S₁(1 + 8S₂)