Question

An arithmetic sequence is defined by the general term tn = 19 - (n - 1)18, where n ∈N and n ≥ 1. What is the recursive formula of the sequence?

Answer 1

`t_n=19-18(n-1)`

`t_(n-1)=19-18(n-2)`

`\implies t_n-t_(n-1)=-18(n-1)+18(n-2)`

`\implies t_n=t_(n-1)-18n+18+18n-36`

`\implies t_n=t_(n-1)-18`

So the recursive formula is


`\begin{cases}t_1=19\\t_n=t_(n-1)-18&\text{for }n>1\end{cases}`

Answer 2

`t_1= \text{19 for n =1}`

`t_n=t_(n-1)-18 \text{ for n }> 1`

Explanation:

Given : General term of arithmetic sequence :
`t_n=19-18(n-1)`

To Find: the recursive formula of the sequence

Solution :

`t_n=19-18(n-1)`

To find the recursive formula of the sequence .

`t_(n-1)=19-18(n-1-1)`

`t_(n-1)=19-18(n-2)`

So, the recursive formula is :

`t_n-t_(n-1)=19-18(n-1)-[19-18(n-2)]`

`t_n-t_(n-1)=19-18n+18-{19-18n+36]`

`t_n-t_(n-1)=19-18n+18-19+18n-36`

`t_n-t_(n-1)=-18`

`t_n=t_(n-1)-18`

Hence the recursive formula of the sequence is :

`t_1= \text{19 for n =1}`

`t_n=t_(n-1)-18 \text{ for n }> 1`