How do you find a_6 for the geometric sequence 540, 90, 15,...?

Oct 23, 2016

The sixth term ${a}_{6}$ is $\frac{5}{72}$

Explanation:

In a geometric series, whose first term is ${a}_{1}$ and common ratio is $r$, ${n}^{t h}$ term ${a}_{n}$ is given by ${a}_{1} \times {r}^{n - 1}$.

In the given series $\left\{540 , 90 , 15 , \ldots .\right\}$, first term is $540$ and common ration $r$ is $\frac{90}{540} = \frac{15}{90} = \frac{1}{6}$

Hence the sixth term ${a}_{6}$ is ${a}_{1} \times {r}^{n - 1} = 540 \times \frac{1}{6} ^ \left(6 - 1\right)$

= $\frac{540}{6 \times 6 \times 6 \times 6 \times 6} = \frac{6 \times 6 \times 3 \times 5}{6 \times 6 \times 6 \times 6 \times 6}$

= $\frac{\cancel{6} \times \cancel{6} \times \cancel{3} \times 5}{\cancel{6} \times \cancel{6} \times 2 \cancel{6} \times 6 \times 6}$

= $\frac{5}{72}$