Question

The first term of the sequence is 8 and the common difference is 12
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Answer 1

$a_n=a_{n-1}+d\\\\a_1=8,\ d=12\\a_2=a_1+d\to a_2=8+12=20\\a_3=a_2+d\to a_3=20+12=32\\a_4=a_3+d\to a_4=32+12=44\\a_5=a_4+d\to a_5=44+12=56\\a_6=a_5+d\to a_6=56+12=68\\\vdots$