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)$\frac{4}{5},-\frac{8}{11},\frac{16}{17},-\frac{32}{23},\frac{64}{29},\dots$

(a)

(b)

(c)

trovabile4p

Beginner2022-01-25Added 13 answers

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${a}_{n}=a+(n-1)d$

$80+(n+1)(-6)$

$80-6n+6=86-6n$ - general term, where n is positive integers

1)

Hence, we have

Palandriy0

Beginner2022-01-26Added 14 answers

(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$\frac{{(-1)}^{n-1}{\left(2\right)}^{n+1}}{6n-1}$

Where n is positive integers

Here

Thus, the above terms are in AP

With first term

We get

(c)

The general form is

Where n is positive integers