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Find the first six terms of the sequence. a1 = 4, an = an - 1 + 8
Solution:
Given, a1 = 4
an = an - 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, an = an - 1 + 8
Summary:
The first six terms of the sequence a1 = 4, an = an - 1 + 8 are 4, 12, 20, 28, 36 and 44.
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