Question

Find the first term, a1, of an arithmetic sequence if a15 = 136 and a19 = 168.

Answer 1

Since $a_n$ is an arithmetic sequence, there is some constant $d$ by which consecutive terms differ. This means

$a_{16}=a_{15}+d$
$a_{17}=a_{16}+d=a_{15}+2d$
$a_{18}=a_{17}+d=a_{15}+3d$
$a_{19}=a_{18}+d=a_{15}+4d$

$168=136+4d\implies d=8$

By a similar token, we can express $a_{15}$ in terms of $a_1$:

$a_2=a_1+d$
$a_3=a_2+d=a_1+2d$
$\cdots$
$a_{15}=a_{14}+d=a_1+14d$

$136=a_1+14\cdot8\implies a_1=24$