Question

find the next three terms of 27,9,3,1,1/3,1/6​

Answer 1

Answer:

$\frac{1}{27}, \frac{1}{81}, \frac{1}{243}$

Step-by-step explanation:

you divide every next term by 3

Answer 2

ANSWER:  

The next three terms of given sequence are  $\frac{1}{27}, \frac{1}{81}, \frac{1}{243}$

SOLUTION:

Given, series is  $27,9,3,1, \frac{1}{3}, \frac{1}{6}$

Let us find which sequence does this series belongs.

$\text { Here } \mathrm{t}_{1}=27, \mathrm{t}_{2}=9, \mathrm{t}_{3}=3$

Now, let us see the common difference.

$t_{2}-t_{1}=9-27=-18$

$\begin{array}{l}{t_{3}-t_{2}=3-9=-6} \\ {t_{2}-t_{1} \neq t_{3}-t_{2}}\end{array}$

So, the series does not belong to arithmetic progression.

Now, let us see the common ratio.

$\begin{array}{l}{\frac{t_{2}}{t_{1}}=\frac{9}{27}=\frac{1}{3}} \\\\ {\frac{t_{3}}{t_{2}}=\frac{3}{9}=\frac{1}{3}} \\\\ {\frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{2}}}\end{array}$

So, the series belongs to geometric progression. And common ratio(r) = $\frac{1}{3}$

We need to find the next three terms, which are $\mathrm{t}_{7}, \mathrm{t}_{8}, \mathrm{t}_{9}$$\text { We know that, } \mathrm{t}_{\mathrm{n}}=\mathrm{t}_{1} \cdot \mathrm{r}^{\mathrm{n}-1}$

$\text { Then, } \mathrm{t}_{7}=27 \times\left(\frac{1}{3}\right)^{7-1} = 27 \times\left(\frac{1}{3}\right)^{6}=27 \times \frac{1}{27 \times 27}=\frac{1}{27}$

$\mathrm{t}_{8}=27 \times\left(\frac{1}{3}\right)^{8-1} = 27 \times\left(\frac{1}{3}\right)^{7} = =27 \times \frac{1}{27 \times 81}=\frac{1}{81}$

$t_{9}=27 \times\left(\frac{1}{3}\right)^{9-1} = 27 \times\left(\frac{1}{3}\right)^{8} = 27 \times \frac{1}{27 \times 243}=\frac{1}{243}$

Hence the next three terms of given sequence are  $\frac{1}{27}, \frac{1}{81}, \frac{1}{243}$