Question

Find the first six terms of the sequence.

Answer 1

The arithmetic sequence:
a 1 = 4,  a n = a n-1 + 8,  d = 8
a 2 = 4 + 8 = 12
a 3 = 12 + 8 = 20
a 4 = 20 + 8 = 28
a 5 = 28 + 8 = 36
a 6 = 36 + 8 = 44
Answer:
C ) 4, 12, 20, 28, 36, 44

Answer 2

Answer:

(C) 4, 12, 20, 28, 36, 44

Step-by-step explanation:

Given: $a_{1}=4$ and $a_{n}=a_{n-1}+8$

To find: Find the first six terms of the sequence

Solution: Since, it is given that $a_{1}=4$ and $a_{n}=a_{n-1}+8$, then the first term is 4 and the common difference, d=8.

Thus, the next terms are:

$a_{2}=a_{1}+8=4+8=12$,

$a_{3}=a_{2}+8=12+8=20$,

$a_{4}=a_{3}+8=10+8=28$,

$a_{5}=a_{4}+8=28+8=36$ and

$a_{6}=a_{5}+8=36+8=44$

Thus, the first six terms of the sequence will be 4, 12, 20, 28, 36, 44.

thus, option C is correct.