Math, asked by Anonymous, 7 months ago

If the sum of first n,2n and 3n terms of an AP be S¹,S²and S³ respectively, then prove that S³=3(S²-S¹)

Answers

Answered by Anonymous

23

Given that

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

S2 = 2×n/2[2a+(2n-1)d]

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

So,LHS, S2-1

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

= 4an +4n2d-2nd-2an-n2d+nd

2

2an+3n2d-nd = n (2a+3nd)

2 2

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

2 2

=3(S1, S2)

HENCE S3 =3( S-S1)

HENCE PROVED.

Answered by taxii

4

Step-by-step explanation:

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