Find the first six terms of the sequence. a₁ = 4, a_n = a_{n - 1} + 8

Solution:

Given, a₁ = 4

a_n = a_{n - 1} + 8

The given sequence is an arithmetic sequence.

We have to find the first six terms of the sequence.

\(\\a_{2}=a_{(2-1)}+8\\a_{2}=a_{1}+8\\a_{2}=4+8\\a_{2}=12\)

\(\\a_{3}=a_{(3-1)}+8\\a_{3}=a_{2}+8\\a_{3}=12+8\\a_{3}=20\)

\(\\a_{4}=a_{(4-1)}+8\\a_{4}=a_{3}+8\\a_{4}=20+8\\a_{4}=28\)

\(\\a_{5}=a_{(5-1)}+8\\a_{5}=a_{4}+8\\a_{5}=28+8\\a_{5}=36\)

\(\\a_{6}=a_{(6-1)}+8\\a_{6}=a_{5}+8\\a_{6}=36+8\\a_{6}=44\)

Therefore, the first six terms of the sequence are 4, 12, 20, 28, 36 and 44.

Find the first six terms of the sequence. a₁ = 4, a_n = a_{n - 1} + 8

Summary:

The first six terms of the sequence a1 = 4, a_n = a_{n - 1} + 8 are 4, 12, 20, 28, 36 and 44.