Question

find the first,fourth,and tenth terms of the arithmetic sequence described by the given rule. A(n)=5+(n-1)(1/6)

Answer 1

a₁ = 5 an = a₁ + ( n-1) * r
wyliczymy jednak - so we enumerate
a₁ = 5 + (1-1) * (1/6) = 5 + 0* 1/6 = 5 + 0 = 5
a₄ = 5 + (4 - 1) * (1/6) = 5 + 3 * (1/6) = 5 + 3/6 =5 1/2 = 5,5
a₁₀ = 5 + (10 - 1) * (1/6) = 5 + 9 * (1/6) = 5 + (9/6) = 5 + 3/2 = 6 1/2 = 6,5

Answer 2

First term is 5

Fourth term is 5.5

Tenth term is 6.5

Explanation:

Given :The rule as
`A(n)=5+(n-1)((1)/(6))`

We have to find find the first,fourth,and tenth terms of the arithmetic sequence.

Consider the given rule ,

`A(n)=5+(n-1)((1)/(6))`

For first term , put n = 1

`A(1)=5+(1-1)((1)/(6))`

Simplify , we have,

`A(1)=5`

For fourth term, Put n = 4

we have,

`A(4)=5+(4-1)((1)/(6))`

Simplify, we have,

`A(4)=5+(3)((1)/(6))`

`A(4)=5+(1)/(2)=5.5`

For tenth term, put n = 10 ,

We have,

`A(10)=5+(10-1)((1)/(6))`

`A(10)=5+(9)((1)/(6))`

`A(4)=5+(3)/(2)=6.5`

Thus, First term is 5

Fourth term is 5.5

Tenth term is 6.5