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The sum of the first n terms of the arithmetic progression is equal to half the sum of the next n terms ofthe same progression. Find the ratio of the sum of the first 3n terms of the progression to the sum of itsfirst n terms.
Dear Vinayak,we have..S(n)=1/2 [ S(2n) - S(n) ]hence 3/2.S(n) = S(2n)put the formula for summation and we get the relation as... 2a=d(1-5n)now put the value of 2a in S(3n)/S(n)....and then you get the ans... As 3/2Regards.
Dear Vinayak, we have.. S(n)=1/2 [ S(2n) - S(n) ] hence 3/2.S(n) = S(2n) put the formula for summation and we get the relation as... 2a=d(1-5n) now put the value of 2a in S(3n)/S(n)....and then you get the ans... As 3/2 Regards.
3/2
sn=sum to n terms
sn/2=s2n-sn
apply sn=n(2a+(n-1)d)
and for s2n replace n by 2n
we get condition 2a=d(1-5n)
then find s3n/sn
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