Let `S_n` denote the sum of first n terms of an AP and `3S_n=S_(2n)` What is `S_(3n):S_n` equal to? What is `S_(3n):S_(2n)` equal to?

Options

A. 4
B. 6
C. 8
D. 10

Answer

B

Explanation

Let first term be a and copmmon difference be d.
then `s_(n)=(n)/(2) [2a+(n-1)d]` ...(i)
` therefore S_(2n) =(2n)/(2) [2a+(2n-1)d]`
` S_(2n) =n[2a+(2n-1)d]` ...(ii)
` S_(3n)=(3n)/(2)[2a+(3n-1)d]`...(iii)
According to the question ,`S_(2n)=3s_(n)`
`implies n[2a+(2n-1)d]=3(n)/(2)[2a+(n-1)d]`
` implies 4a+(4n-2)d=6a+(3n-3)d`
`implies -2a+(4n-2-3n+3)d=0 `
`implies d=(2a)/(n+1)`
Now`(S_(3n))/(S_(n))=((3n)/(2)[a+(3n-1)d])/((n)/(2)[2a+(n-1)d])=(6a+(9n-3)(2a)/(n+1))/(2a+(n+1)(2a)/(n+1)) `
`=(6an+6a+18an-6a)/(2an+2a+2an-2a)`
`=(24an)/(4an )/(S_(3n))/(s_(n))=6`