1.

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?

Answer» `S_n=n/2[2a+(n-1)d]`
`3S_n=S_(2n)`
`3(n/2)[2a+(n-1)d]=(2n)/2[2a+(2n-1)d]`
`3/2[2a+(n-1)d]=2a+(2n-1)d`
`3a+3/2nd-3/2d=2a+2nd-d`
`a=(nd)/2+d/2`
`a=(n+1)d/2`
`2a=(n+1)d`
`S_(3n)/S_n=((3n/2)[2a+(3n-1)d])/(n/2[2a+(n-1)d]`
`=(3[nd+d+3nd-d])/(nd+d+nd-d)`
`(12nd)/(2nd)`
`S_(3n):S_n=6:1`
option b is correcct.


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