Given:
[tex]a_n=46+6(n-1)[/tex]Find: First and forth term of the sequence.
Explanation:
[tex]\begin{gathered} a_n=46+6(n-1) \\ n=1 \\ a_1=46+6(1-1) \\ =46 \\ n=2 \\ a_2=46+6(2-1) \\ =46+6 \\ =52 \\ n=3 \\ a_3=46+6(3-1) \\ =46+6(2) \\ =46+12 \\ =58 \\ n=4 \\ a_{4=}=46+6(4-1) \\ =46+6(3) \\ =46+18 \\ =64 \end{gathered}[/tex]Final answer: the first four term of the sequence are 46,52,58 and 64
