The sum of n,2n,3n terms of an AP are s1,s2,s3 respectively. Then prov

1.	The sum of n,2n,3n terms of an AP are s1,s2,s3 respectively. Then prove that s3={s2-s1}3 .
Answer» S1 = n/2 (2a + (n-1)d) ;S2 = n(2a + (2n-1)d) ;S3 = 3n/2(2a+(3n/2-1)d) ;3(S2 -S1)= 3( 2an +2n^2d -dn -an -n^2d/2 +nd/2)= 3( an +3dn^2 /2 -nd/2)= 3n/2 (2a + (3n/2 -1) d)= S3Hence, proved

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