Kayleigh Dixon

2022-01-24

Given the first 5 terms, write down the general term
(a) $80,74,68,62,56\dots$
(b)$3,1,\frac{1}{3},\frac{1}{9},\frac{1}{27},\dots$
(c) $\frac{4}{5},-\frac{8}{11},\frac{16}{17},-\frac{32}{23},\frac{64}{29},\dots$

trovabile4p

We are given the first terms and we have to find the general form
1)$80,74,68,62,56\dots$
${a}_{2}-a={a}_{3}-{a}_{2}={a}_{4}-{a}_{3}={a}_{5}-{a}_{4}$
$74-80=687-74=62-68=56-62$
$-6=-6=-6=-6$
Hence, we have ${a}_{n}=a+\left(n-1\right)d$
$80+\left(n+1\right)\left(-6\right)$
$80-6n+6=86-6n$ - general term, where n is positive integers

Palandriy0

(b)$3,1,\frac{1}{3},\frac{1}{9},\frac{1}{27},\dots$
Here $d=\frac{1}{3}$
Thus, the above terms are in AP
With first term $=3$ and common ratio $=\frac{1}{3}$
We get $3{\left(\frac{1}{3}\right)}^{n-1}=9{\left(\frac{1}{3}\right)}^{n}$ is the general form
(c) $\frac{4}{5},-\frac{8}{11},\frac{16}{17},-\frac{32}{23},\frac{64}{29},\dots$
The general form is $\frac{{\left(-1\right)}^{n-1}{\left(2\right)}^{n+1}}{6n-1}$
Where n is positive integers

Do you have a similar question?

Recalculate according to your conditions!