The fourth, seventh, and the last term of a G.P. are 10, 80, and 2560, respectively. Find the first term and the number of terms in G.P.

Explanation

Let a be the first term and r be the common ratio of the given G.P. Then,
`a_(4)=10,a_(7)=80`
`rArrar^(3)=10andar^(6)=80`
`rArr(ar^(6))/(ar^(3))=80/10or r^(3)=8orr=2`
Putting r=2 in `ar^(3)=10`, we get a=10/8.
Let there be n terms in the given G.P. Then,
`a_(n)=2560orar^(n-1)=2560`
or `10/8(2^(n-1))=2560`
or `2^(n-4)=256or2^(n-4)=2^(8)`
or n-4=8 or n=12