Question

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

Answer 1

Answer with Step-by-step explanation:

We are given a arithmetic sequence as:

$A(n)=-6+(n-1)(\dfrac{1}{5})$

We have to find the first, fourth and tenth term

First term:

n=1

$A(1)=-6+(1-1)(\dfrac{1}{5})$

A(1)= -6

Fourth term:

n=4

$A(4)=-6+(4-1)(\dfrac{1}{5})$

$A(4)=-6+\dfrac{3}{5}$

$A(4)=-\dfrac{27}{5}$

Tenth term:

n=10

$A(10)=-6+(10-1)(\dfrac{1}{5})$

$A(10)=-6+\dfrac{9}{5}$

$A(10)=-\dfrac{21}{5}$

Answer 2

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