Question

The sums of the first n,2n and 3n terms of an arithmetic progression are S1,S2 and S3.Prove that S3=3(S2-S1).

Answer 1

ANSWER:-

Given:

The sums of the first n, 2n & 3n terms of A.P. are S1, S2 & S3.

To prove:

Prove that S3= 3(S2 -S1).

Proof:

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

We know that Formula of the sum of A.P.;

$= > {}^{s} n = \frac{n}{2} [2a + (n - 1)d]$

According to the question:

$s1 = \frac{n}{2} [2a + (n - 1)d]..........(1) \\ \\ s2 = \frac{n}{2} [2a + (n - 1)d].........(2) \\ \\ s3 = \frac{n}{2} [2a + (n - 1)d]..........(3)$

Take R.H.S

3(S2-S1)

$= > 3( \frac{2n}{2} [2a + (2n - 1)d] - \frac{n}{2} [2a + (n - 1)d)]\\ \\ = > 3 \times \frac{n}{2} [4a + (4n - 2)d - 2a - (n - 1)d] \\ \\ = > \frac{3n}{2} [2a + (4n - 2 - n + 1)d] \\ \\ = > \frac{3n}{2} [2a + (3n - 1)d] = s3 \: \: \: \: \: \: \: [L.H.S.]$

Hence,

S3 = 3(S2 - S1)

Proved.

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