The sum of n 2n 3n terms of an AP are S1 S2 S3 respectively Prove that S3=3(S2-S1) - Maths - Arithmetic Progressions

Question

The sum of n 2n 3n terms of an AP are S1 S2 S3 respectively
Prove that S3=3(S2-S1)

Mayank Bhardwaj · Accepted Answer

Dear Student!

Here is the answer to your question.

Let the first term of the A.P. be a and the common difference be d.

According the question,

$S_{1} = \frac{n}{2} {2 a + (n - 1) d} ..... (1) S_{2} = \frac{2 n}{2} {2 a + (2 n - 1) d} ..... (2) S_{3} = \frac{3 n}{2} {2 a + (3 n - 1) d} ..... (3)$

We have to prove that S₃ = 3 ( S₂ – S₁).

R.H.S = 3 (S₂ – S₁)

$= 3 [\frac{2 n}{2} {2 a + (2 n - 1) d} - \frac{n}{2} {2 a + (n - 1) d}] = 3 \times \frac{n}{2} {4 a + (4 n - 2) d - 2 a - (n - 1) d} = \frac{3 n}{2} {2 a + (4 n - 2 - n + 1) d} = \frac{3 n}{2} {2 a + (3 n - 1) d}$

= S₃

= L.H.S

⇒ S₃ = 3 ( S₂ – S₁)

Cheers !

The sum of n 2n 3n terms of an AP are S1 S2 S3 respectively. Prove that S3=3(S2-S1).