Question

What is the 9th term of the geometric sequence where a1 = 625 and a3 = 25?

Answer 1

`a_9=1/625`

Explanation:

Let the geometric series be `{a_1,a_2,a_3,a_4,.....}`, where `(a_2)/(a_1)=(a_3)/(a_2)=(a_4)/(a_3)=(a_5)/(a_4)=.....=r`.

the `n^(th)` of such a series is given by `a_n=axxr^(n-1)`

As `a_1=625`, hence `a_3=625xxr^2`, but this is `25`.

Hence `625xxr^2=25` or `r^2=1/25` or `r=+-1/5`

Hence `9^(th)` term `a_9=625xx(+-1/5^8)`

or `a_9=625/25^4=1/625`