Question

Find the first six terms of the sequence.

a1 = -4, an = an-1 + 7

Answer 1

`a_n=a_(n-1)+7\\\\a_1=-4\\a_2=-4+7=3\\a_3=3+7=10\\a_4=10+7=17\\a_5=17+7=24\\a_6=24+7=31`

Answer 2

first six terms are -4, 3, 10, 17, 24 ,31

Explanation:

The given first term and recurrence relation is
`a_1 = -4, \; a_n = a_(n-1) +7`

we need to find first six term of this recurrence relation

`a_1 = -4`

put n=2 in
`a_n = a_(n-1) +7`

`a_2 = a_(2-1) +7`

`a_2 = a_1 +7`

since,
`a_1 = -4`

`a_2 = -4 +7`

`a_2 = 3`

similarly,

for n=3

`a_3 = a_(3-1) +7`

`a_3 = a_2 +7`

since,
`a_2 = 3`

`a_3 = 3 +7`

`a_3 = 10`

For n=4

`a_4 = a_(4-1) +7`

`a_4 = a_3 +7`

since,
`a_3 = 10`

`a_4 = 10 +7`

`a_4 = 17`

For n=5

`a_5 = a_(5-1) +7`

`a_5 = a_4 +7`

since,
`a_4 = 17`

`a_5 = 17 +7`

`a_5 = 24`

For n=6

`a_6 = a_(6-1) +7`

`a_6 = a_5 +7`

since,
`a_5 = 24`

`a_6 = 24 +7`

`a_6 = 31`

Hence, first six terms are -4, 3, 10, 17, 24 ,31