Question

Find the first six terms of the sequence. (1 point)

a1 = 7, an = an-1 + 6

a) 13, 19, 25, 31, 37, 43

b) 0, 6, 12, 18, 24, 30

c) 7, 6, 12, 18, 24, 30

d) 7, 13, 19, 25, 31, 37

Answer 1

Option d - 7, 13, 19, 25, 31, 37

Explanation:

Given :
`a_1=7` and
`a_n=a_(n-1)+6`

To find : The first six terms of the sequence?

Solution :

We have given the nth formula,

`a_n=a_(n-1)+6`

The first term is
`a_1=7`

For second term put n=2,

`a_2=a_(2-1)+6`

`a_2=a_(1)+6`

`a_2=7+6`

`a_2=13`

For third term put n=3,

`a_3=a_(3-1)+6`

`a_3=a_(2)+6`

`a_3=13+6`

`a_3=19`

For fourth term put n=4,

`a_4=a_(4-1)+6`

`a_4=a_(3)+6`

`a_4=19+6`

`a_4=25`

For fifth term put n=5,

`a_5=a_(5-1)+6`

`a_5=a_(4)+6`

`a_5=25+6`

`a_5=31`

For sixth term put n=6,

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

`a_6=a_(5)+6`

`a_6=31+6`

`a_6=37`

Therefore, The required sequence is 7, 13, 19, 25, 31, 37.

So, Option d is correct.

Answer 2

by the given data above, we can tell that the arithmetic difference of the sequence is 6 as the difference between an and an-1 is so. SO we start the sequence

a1 = 7
a2 = 13
a3 = 19
a4 = 25
a5 = 31
a6 = 37

Answer to this problem is D