Question

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

Answer 1

Answer:

The first term is $A(1)=5$

The fourth term is $A(4)=\frac{11}{2}$.

The tenth term is $A(10)=\frac{13}{2}$

Step-by-step explanation:

Given : Arithmetic sequence  $A(n)=5+(n-1)( \frac{1}{6} )$

To find : The first, fourth, and tenth terms of the arithmetic sequence described by the given rule?

Solution :

Arithmetic sequence  $A(n)=5+(n-1)(\frac{1}{6})$

Substitute, n=1,4,10 to find the given terms

Put n=1,

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

$A(1)=5+(0)(\frac{1}{6})$

$A(1)=5$

The first term is $A(1)=5$

Put n=4,

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

$A(4)=5+(3)(\frac{1}{6})$

$A(4)=5+\frac{1}{2}$

$A(4)=\frac{11}{2}$

The fourth term is $A(4)=\frac{11}{2}$.

Put n=10,

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

$A(10)=5+(9)(\frac{1}{6})$

$A(10)=5+\frac{3}{2}$

$A(10)=\frac{13}{2}$

The tenth term is $A(10)=\frac{13}{2}$

Answer 2

$A(1)=5+(1-1)(\frac{1}{6})=5+0(\frac{1}{6})=5+0=5$
$A(14)=5+(5-1)(\frac{1}{6})=5+4(\frac{1}{6})=5+\frac{4}{6}=5\frac{2}{3}$
$A(10)=5+(10-1)(\frac{1}{6})=5+9(\frac{1}{6})=5+\frac{9}{6}=6\frac{1}{2}$