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

see explanation

Explanation:

The common difference d of an arithmetic sequence is

d =
`a_(2)` -
`a_(1)` =
`a_(3)` -
`a_(2)`

Substitute in values and solve for k, that is

5k - 1 - 2k = 6k + 2 - (5k - 1)

3k - 1 = 6k + 2 - 5k + 1

3k - 1 = k + 3 ( subtract k from both sides )

2k - 1 = 3 ( add 1 to both sides )

2k = 4 ⇒ k = 2

--------------------------------------------------------

The n th term of an arithmetic sequence is

`a_(n)` =
`a_(1)` + (n - 1)d

`a_(1)` = 2k = 2 × 2 = 4 and

d = 5k - 1 - 2k = 3k - 1 = (3 × 2) - 1 = 5

Hence

`a_(8)` = 4 + (7 × 5) = 4 + 35 = 39