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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 of the same progression. Find the ratio of the sum of the first 3n terms of the progression to the sum of its first n terms.

The sum of the first n terms of the arithmetic progression is equal to half the sum of the next n terms of
the same progression. Find the ratio of the sum of the first 3n terms of the progression to the sum of its
first n terms.

Grade:

4 Answers

Akash Kumar Dutta
98 Points
8 years ago

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.

Akash Kumar Dutta
98 Points
8 years ago

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.

konduri abhilash reddy reddy
18 Points
8 years ago

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

mnjay
20 Points
6 years ago
right ans is 6
if we solve 
S2N/SN=3
it gives relation as 2a=(n+1)d
now take ratio of S3N/SN and put 2a=(n+1)d you will get 6

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