Determine the number of terms in G.P. `<>,ifa_1=3,a_n=96a n dS_n=189.`

Explanation

Let r be the common ratio of the given G.P. Then
`a_(n)=96`
or `a_(1)r^(n-1)=96`
or `3r^(n-1)`=96
or `r^(n-1)=32`
Now, `S_(n)=189`
or `a_(1)((r^(n)-1)/(r-1))=189`
or`3(((r^(n-1))r-1)/(r-1))=189`
or `3((32r-1)/(r-1))=189`
or 32r-1=63r-63
or 31r=62
or r=2
Putting r=2 in (1), we get
`2^(n-1)=32`
or `2^(n-1)=2^(5)`
or n-1=5
or n=6