Question

if the sum of first n,2n,3n terms of an a.p are s1 ,s2,s3 respectively ,show that s3 =3(s2-s1)

Answer 1

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Answer 2

S1, S2, S3 are sums of first n, 2n, 3n terms

to prove: S3 = 3 ( S2 - S1 )

proof:
S1 = Sn = (2a + (n-1)d)n/2
S2 = S2n = (2a + (2n-1)d)2n/2
S3 = S3n = (2a + (3n-1)d)3n/2 .... ( i )

RHS: 3 ( S2 - S1 )
= 3 { [ (2a + (2n-1)d)2n/2] - [ (2a + (n-1)d)n/2] }
= 3n/2 [2( 2a + (2n-1)d) - (2a + (n-1)d)]
= 3n/2 ( 4a + 4nd - 2d - 2a - nd + d )
= 3n/2 ( 2a + 3nd - d )
= (2a+(3n-1)d)3n/2
= S3n = S3
= LHS
hence proved

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