Respuesta :
[tex]\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}[/tex]
Answer with Step-by-step explanation:
We are given a arithmetic sequence as:
[tex]A(n)=-6+(n-1)(\dfrac{1}{5})[/tex]
We have to find the first, fourth and tenth term
First term:
n=1
[tex]A(1)=-6+(1-1)(\dfrac{1}{5})[/tex]
A(1)= -6
Fourth term:
n=4
[tex]A(4)=-6+(4-1)(\dfrac{1}{5})[/tex]
[tex]A(4)=-6+\dfrac{3}{5}[/tex]
[tex]A(4)=-\dfrac{27}{5}[/tex]
Tenth term:
n=10
[tex]A(10)=-6+(10-1)(\dfrac{1}{5})[/tex]
[tex]A(10)=-6+\dfrac{9}{5}[/tex]
[tex]A(10)=-\dfrac{21}{5}[/tex]
