Question

find the first,fourth,and tenth terms of the arithmetic sequence described by the given rule. A(n)=5+(n-1)(1/6)

Answer 1

A₁ = 5    an = a₁ + ( n-1) * r
wyliczymy jednak - so we enumerate
a₁ = 5 + (1-1) * (1/6) = 5 + 0* 1/6 = 5 + 0 = 5
a₄ = 5 + (4 - 1) * (1/6) = 5 + 3 * (1/6) = 5 + 3/6 =5 1/2 = 5,5
a₁₀ = 5 + (10 - 1) * (1/6) = 5 + 9 * (1/6)  =  5 + (9/6) = 5 + 3/2 = 6 1/2 = 6,5

Answer 2

Answer:

First term is 5

Fourth term is 5.5

Tenth term is 6.5                  

Step-by-step explanation:

Given :The rule as $A(n)=5+(n-1)(\frac{1}{6})$        

We have to find  find the first,fourth,and tenth terms of the arithmetic sequence.

Consider the given rule ,

$A(n)=5+(n-1)(\frac{1}{6})$        

For first term , put n = 1

$A(1)=5+(1-1)(\frac{1}{6})$    

Simplify , we have,  

$A(1)=5$            

For fourth term, Put n = 4

we have,

$A(4)=5+(4-1)(\frac{1}{6})$    

Simplify, we have,

$A(4)=5+(3)(\frac{1}{6})$    

$A(4)=5+\frac{1}{2}=5.5$  

For tenth term, put n = 10 ,

We have,

$A(10)=5+(10-1)(\frac{1}{6})$    

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

$A(4)=5+\frac{3}{2}=6.5$  

Thus, First term is 5

Fourth term is 5.5

Tenth term is 6.5