Find the sum to indicated number of terms in each of the geometric progressions in Questions 7 to 10 :
`1,-a,a^(2),-a^(3),..."n terms "(ifa ne-1)`

Answer

N/a

Explanation

`"Given G.P : "1,-a,a^(2),-a^(3),...(ane-1)`
`"1st term "a_(1)=a=1`
`"2nd term "a_(2)=-a`
`therefore" Common ratio "r=(a_(2))/(a_(1))=(-a)/(1)=-a`
Then, sum of n terms of G.P.
`S_(n)=(a(1-r^(n)))/((1-r))`
`=(1[1-(-a)^(n)])/([1-(-a)])=([1-(-1)^(n)cdota^(n)])/((1+a))`
`therefore" Sum of n terms of given G.P. is "S_(n)=(1-(-1)^(n)a^(n))/(1+_a).`