Question

sum of n,2n,3n, term of an AP are S1,S2,and S3,.

PROVE That S3=3(S2-S1).

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Answer 1

Here is your answer :-

A.T.Q.

S1 = n/2[2a+(n-1)d] -------(1)

S2 = 2n/2[2a+(2n-1)d]
= n [2a+(2n-1)d] -------(2)

S3 = 3n/2[2a+(3n-1)d] --------(3)

S3 =3(S2-S1)

3n/2[2a+(3n-1)d] = 3(2an+2n²d-nd-(2an+2n²d-nd)/2)


3n/2[2a+(3n-1)d] = 3/2(4an+4n²d-2nd-2an-2n²d+nd)


3n/2[2a+(3n-1)d] = 3/2(2an +3n²d -nd)


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

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

Hence proved ...

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