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The sum of n, 2n , 3n terms of an AP are S1,S2 ,S3 respectively. Prove that S3 = 3(S2 -S1).

Posted by Nancy Dewangan 6 years, 2 months ago

CBSE > Class 10 > Mathematics

3 answers

:) ** 6 years, 2 months ago

S1= n/2 [2a+(n-1) d] S2= 2n/2 [2a+(2n-1) d] S3= 3n/2 [2a+(3n-1) d] .......(i) S2-S1 = 2n/2 [2a +(2n-1) d] - n/2 [2a+(n-1) d ] = n/2[2a +(2n-1) 2d - 2a -(n-1) d ] = n/2 [2a + ( 4n-2-n+1) d ] = n/2 [ 2a + (3n-1) d ] (ie) 3 (S2-S1) = 3n/2 from (i) Hence proved .

1Thank You

Sri Varma 6 years, 2 months ago

S1=a1 S2 =a1+a2=n+2n=3n S3 =a1+a2+a3=n+2n+3n=6n Lhs=S3=6n Rhs=3(3n-n) Rhs=6n Lhs=Rhs Hp

1Thank You

Madhwinder Bajwa 6 years, 2 months ago

It is not possible

1Thank You

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CBSE > Class 10 > Mathematics