The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Find the sum of `n` terms of the G.P.

Answer

N/a

Explanation

Let first term be 'a' and common ratio 'r' Now `S_(3)=1`
`implies (a(1-r^(3)))/(1-r)=16`….(1)
`and S_(6)=16+128`
`implies (a(1-r^(6)))/(1-r)=144`
`implies (a(1-r^(3)))/(1-r)*(1+r^(3))=144`
`implies 16*(1+r^(3))=144`
`implies 1+r^(3)=9`
`impliesr^(3)=8`
`impliesr=2`
Now `(a(1-r^(3)))/(1-r)=16` [Put r=2 in equation (1)]
`implies (a(1-8))/(1-2)=16`
`implies a=(16)/(7)`
and `S_(n)=(a(r^(n)-1))/(r-1)`
`S_(n)=((16)/(7)(2^(n)-1))/(2-1)=(16)/(7)(2^(n)-1)`
`:. a=(16)/(7),r=2,S_(n)=(16)/(7)(2^(n)-1)`