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
![a_n=a_1+d(n-1)](https://img.qammunity.org/2023/formulas/mathematics/college/z9jyjzs3gtye2ac99tljhz0dstsxu67bsc.png)
we have
a1=14
d=6
substitute
![\begin{gathered} a_n=14+6(n-1) \\ a_n=14+6n-6 \\ a_n=8+6n \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/560xlpjnk7emt3yfhsz8fwlfmwvpr5pay6.png)
find the 63rd term
For n=63
substitute in the formula
![\begin{gathered} a_n=8+6(63) \\ a_n=386 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/46b2on59yv9pxij51zdiwm19ji16wft2gi.png)
the answer is 386