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 3/2
Regards.

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