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`