Question

Find the first, fourth, and 10th terms of the arithmetic sequence described by the given rule.

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

Answer 1

`\bf \begin{array}{ll} \stackrel{term}{n}&\stackrel{-6+(n-1)(1)/(5)}{value}\\ \cline{1-2} 1&-6+(1-1)(1)/(5)\\ &-6+0\\[1em] &-6\\[1em] 4&-6+(4-1)(1)/(5)\\ &-6+(3)/(5)\\[1em] &(-27)/(5)\\[1em] 10&-6+(10-1)(1)/(5)\\ &-6+(9)/(5)\\[1em] &(-21)/(5) \end{array}`

Answer 2

Answer with Step-by-step explanation:

We are given a arithmetic sequence as:

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

We have to find the first, fourth and tenth term

First term:

n=1

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

A(1)= -6

Fourth term:

n=4

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

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

`A(4)=-(27)/(5)`

Tenth term:

n=10

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

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

`A(10)=-(21)/(5)`