Question

If the sum of n terms of an AP is Sn and if S2n=3Sn then prove that S3n = 6Sn

Answer 1

GIVEN:

• Sum of n terms of an A.P is Sₙ

• S₂ₙ = 3Sₙ

TO PROVE:

• S₃ₙ = 6Sₙ

PROOF:

Sₙ = n/2(2a + (n - 1)d)

S₂ₙ = 2n/2(2a + (2n - 1)d)

Given that : S₂ₙ = 3Sₙ

2n/2(2a + (2n - 1)d) = 3(n/2(2a + (n - 1)d))

Cancel n/2 on both sides

2(2a + 2nd - d) = 3(2a + nd - d)

4a + 4nd - 2d = 6a + 3nd - 3d

2a - nd - d = 0

Equate : S₃ₙ = 6Sₙ

S₃ₙ = 3n/2(2a + 3nd - d)

Add and subtract by nd

S₃ₙ = 3n/2(2a + 4nd - nd - d)

Substitute 2a - nd - d = 0

S₃ₙ = 3n/2(0 + 4nd)

S₃ₙ = 3n(2nd)

S₃ₙ = 6n(nd)

Sₙ = n/2(2a + nd - d)

Add and subtract by nd

Sₙ = n/2(2a + 2nd - nd - d)

Substitute 2a - nd - d = 0

Sₙ = n/2(0 + 2nd)

Sₙ = n(nd)

6Sₙ = 6(n(nd))

S₃ₙ = 6Sₙ

HENCE PROVED.