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

Mar 9, 2016

${a}_{9} = \frac{1}{625}$

Explanation:

Let the geometric series be $\left\{{a}_{1} , {a}_{2} , {a}_{3} , {a}_{4} , \ldots . .\right\}$, where $\frac{{a}_{2}}{{a}_{1}} = \frac{{a}_{3}}{{a}_{2}} = \frac{{a}_{4}}{{a}_{3}} = \frac{{a}_{5}}{{a}_{4}} = \ldots . . = r$.

the ${n}^{t h}$ of such a series is given by ${a}_{n} = a \times {r}^{n - 1}$

As ${a}_{1} = 625$, hence ${a}_{3} = 625 \times {r}^{2}$, but this is $25$.

Hence $625 \times {r}^{2} = 25$ or ${r}^{2} = \frac{1}{25}$ or $r = \pm \frac{1}{5}$

Hence ${9}^{t h}$ term ${a}_{9} = 625 \times \left(\pm \frac{1}{5} ^ 8\right)$

or ${a}_{9} = \frac{625}{25} ^ 4 = \frac{1}{625}$