Question

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

Answer 1

`(1)/(27), (1)/(81), (1)/(243)`

Explanation:

you divide every next term by 3

Answer 2

The next three terms of given sequence are
`(1)/(27), (1)/(81), (1)/(243)`

SOLUTION:

Given, series is
`27,9,3,1, (1)/(3), (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) \\eq 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}{(t_(2))/(t_(1))=(9)/(27)=(1)/(3)} \\\\ {(t_(3))/(t_(2))=(3)/(9)=(1)/(3)} \\\\ {(t_(2))/(t_(1))=(t_(3))/(t_(2))}\end{array}`

So, the series belongs to geometric progression. And common ratio(r) =
`(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 *\left((1)/(3)\right)^(7-1) = 27 *\left((1)/(3)\right)^(6)=27 * (1)/(27 * 27)=(1)/(27)`

`\mathrm{t}_(8)=27 *\left((1)/(3)\right)^(8-1) = 27 *\left((1)/(3)\right)^(7) = =27 * (1)/(27 * 81)=(1)/(81)`

`t_(9)=27 *\left((1)/(3)\right)^(9-1) = 27 *\left((1)/(3)\right)^(8) = 27 * (1)/(27 * 243)=(1)/(243)`

Hence the next three terms of given sequence are
`(1)/(27), (1)/(81), (1)/(243)`