Question

Find the 64th term of the Arithmetic sequence 29, 38, 47

Answer 1

`a_(64)=596`

Step-by-step explanation

Step 1

find the difference

Let

`\begin{gathered} a_1=29 \\ a_2=38 \\ a_3=47 \end{gathered}`

so, the difference is

`\begin{gathered} a_2-a_1=38-29=9 \\ a_3-a_2=47-38=9 \end{gathered}`

so, the difference is 9, in other words you have to add 9 to the last number to get the new one

Step 2

find the rule

`\begin{gathered} a_1=29 \\ a_2=29+9 \\ a_3=29+9+9 \\ \text{.} \\ \text{.} \\ a_n=29+9\cdot(n-1) \end{gathered}`

Step 3

let

n=64

replace,

`\begin{gathered} a_n=29+9\cdot(n-1) \\ a_(64)=29+9\cdot(64-1) \\ a_(64)=29+9(63) \\ a_(64)=29+567 \\ a_(64)=596 \end{gathered}`

I hope this helps you