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

Step-by-step 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