Question

, list the first five terms of each sequence.
an+1 = an + 6, where a1 = 11 for n ≥ 1

Answer 1

Answer: 17 , 23 , 29 , 35 and 41

Explanation:

Given :
`a_(n+1)` =
`a_(n)` + 6 , where n≥1 , it means n can pick values from 1 and above

`\\`When n = 1 , the sequence becomes

`a_(1+1)` =
`a_(1)` + 6

`\\`And it has been given that
`a_(1)` = 11 , substitute this value into the sequence, that is

`\\`
`a_(2)` = 11 + 6

`\\`
`a_(2)` = 17

`\\`Also , when n = 2 , the sequence becomes

`\\`
`a_(3)` =
`a_(2)` + 6

`\\`substituting the value of
`a_(2)` , we have

`\\`
`a_(3)` = 17 + 6 = 23

`\\`When n = 3 , the sequence becomes

`\\`
`a_(4)` =
`a_(3)` + 6

`\\`substituting the value of
`a_(3)`, we have

`\\`
`a_(4)` = 23 + 6 = 29

`\\`When n = 4 , the sequence becomes

[
`\\`
`a_(5)` =
`a_(4)` + 6 , substituting the value of
`a_(4)` , we have

`\\`
`a_(5)` = 29 + 6 = 35

`\\`And finally , when n = 5 , the sequence becomes

`\\`
`a_(6)` =
`a_(5)` + 6

`\\`Substituting the value of
`a_(5)` , we have

`\\`
`a_(6)` = 35 + 6 = 41

`\\`Therefore , the first five terms are : 17 , 23 , 29 ,35 and 41