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

Answer:

see explanation

Step-by-step 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

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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