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Sagot :
Find the 63 term of the following arithmetic sequence. 14, 20, 26, 32
we have
a1=14
a2=20
a3=26
a4=32
so
a2-a1=20-14=6
a3-a2=26-20=6
a4-a3=32-26=6
therefore
the common difference d=6
The general equation for an arithmetic sequence is equal to
[tex]a_n=a_1+d(n-1)[/tex]we have
a1=14
d=6
substitute
[tex]\begin{gathered} a_n=14+6(n-1) \\ a_n=14+6n-6 \\ a_n=8+6n \end{gathered}[/tex]find the 63rd term
For n=63
substitute in the formula
[tex]\begin{gathered} a_n=8+6(63) \\ a_n=386 \end{gathered}[/tex]the answer is 386
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