Question

If 2k, 5k-1 and 6k+2 are the first 3 terms of an arithmetic sequence, find k and the 8th term.

Answer 1

Consecutive terms in an arithmetic sequence differ by a constant $d$. So

$5k-1=2k+d$
$6k+2=5k-1+d$
$\implies\begin{cases}3k-d=1\\k-d=-3\end{cases}\implies k=2,d=5$

Denote the $n$-th term in the sequence by $a_n$. Now that $a_1=2$, we have

$a_n=a_{n-1}+5=a_{n-2}+2\cdot5=\cdots=a_1+(n-1)\cdot5$

which means

$a_8=2+(8-1)\cdot5=37$