Question

6. Find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule.
A(n)
12 + (n-1)(3)
3,24, 27
0,9,27
12,24, 42
12, 21, 39

Answer 1

Answer:

$A_1 = 12$

$A_4=21$

$A_{10}= 39$

Step-by-step explanation:

$A_n = 12+(n-1)(3) \\ \\ plug \: n = 1 \\ A_1 = 12+(1-1)(3) \\ = 12 + 0(3) \\ = 12 + 0 \\ \red{ A_1= 12} \\ \\ plug \: n = 4\\ A_4 = 12+(4-1)(3) \\ = 12 + 3(3) \\ = 12 + 9 \\ \purple{ A_4= 21} \\ \\ plug \: n = 10\\ A_{10} = 12+(10-1)(3) \\ = 12 + 9(3) \\ = 12 + 27 \\ \orange{ A_{10}= 39}$