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

1 Answer

3 votes

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

User Bubismark
by
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