The 5th and 8th terms of a geometric sequence of real numbers are 7! And 8! Respectively. If the sum to first `n` tems of the G.P. is 2205, then `n` equals_______.

Answer

3

Explanation

Let `a,ar,ar^(2),ar^(3),…` are in G.P.
Now `ar^(4)=7!` and `ar^(7)=8!`
On dividing, we get `r^(3)=8` or r=2
or `a=5040/16=315`
So, 315, 630, 1260,… are in G.P.
`thereforeS_(3)=2205`
`rArrn=3`