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

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

Answer 2

Answer:

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

Step-by-step 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.