Question

the sum of n ,2n,3n terms of an ap is s1, s2,s3 respectively prove that s3=3(s2-s1)

Answer 1

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Answer 2

Answer:

If the sum of n ,2n,3n terms of an AP is s1, s2,s3 , then s3=3(s2-s1)

Step-by-step explanation:

Given: The sum of n ,2n,3n terms of an ap is s1, s2,s3.

To prove: s3=3(s2-s1)

The sum of n, 2n and 3n terms of an AP is

$S_1=S_n=\frac{n}{2}[2a+(n-1)d]$

$S_2=S_{2n}=\frac{2n}{2}[2a+(2n-1)d]$

$S_3=S_{3n}=\frac{3n}{2}[2a+(3n-1)d]$

Simplify R.H.S.

$3(S_2-S_1)=3(\frac{2n}{2}[2a+(2n-1)d]-\frac{n}{2}[2a+(n-1)d])$

$3(S_2-S_1)=3\times \frac{n}{2}\times (2[2a+2dn-d]-[2a+dn-d])$

$3(S_2-S_1)=\frac{3n}{2}(4a+4dn-2d-2a-dn+d)$

$3(S_2-S_1)=\frac{3n}{2}(2a+3dn-d)$

$3(S_2-S_1)=\frac{3n}{2}(2a+(3n-1)d)$

$3(S_2-S_1)=S_3$

Hence proved.