Question

Find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule. A(n)=-6+(n-1)(1/5) A)-6,-5 1/5, -4 B)5, -5 1/5,1 4/5 C)-6, -5 2/5, -4 1/5 D) 0, 3/5, 1 4/5

Answer 1

Step-by-step explanation:

An arithmetic sequence is given by relation as follows :

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

For the first term, put n = 1. So,

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

For fourth term, put n = 4. So,

$A(4)=-6+(4-1)\dfrac{1}{5}\\\\A(4)=\dfrac{-27}{5}\\\\A(4)=-5\dfrac{2}{5}$

For tenth term, put n = 10. So,

$A(10)=-6+(10-1)\dfrac{1}{5}\\\\A(10)=-6+\dfrac{9}{5}\\\\A(10)=\dfrac{-21}{5}\\\\A(10)=-4\dfrac{1}{5}$

Hence, the correct option is (C).

Answer 2

Answer:

the answer in C)-6, -5 2/5, -4 1/5