Question

Consider the given recursive function of an arithmetic sequence.
F(1)=4
F(n)=f(n-1) +7, for n=2,3,4.....

What is the 8th term of the sequence?
A. 60
B. 53
C. 46
D. 67

Answer 1

Method 1:

$F(1)=4\\\\F(n)=F(n-1)+7\\\\n\in\{2,\ 3,\ 4,\ 5,\ ...\}\\\\F(2)=F(1)+7\to F(2)=4+7=11\\\\F(3)=F(2)+7\to F(3)=11+7=18\\\\F(4)=F(3)+7\to F(4)=18+7=25\\\\F(5)=F(4)+7\to F(5)=25+7=32\\\\F(6)=F(5)+7\to F(6)=32+7=39\\\\F(7)=F(6)+7\to F(7)=39+7=46\\\\F(8)=F(7)+7\to F(8)=46+7=53$

Answer: B. 53

Method 2:

The general formula of an arithmetic sequence:

$a_n=a_1+(n-1)d$

We have:

$a_1=F(1)=4,\ d=7$

Substitute:

$a_n=4+(n-1)(7)=4+7n-7=7n-3$

Put n = 8 to the formula:

$a_8=7(8)-3=56-3=53$

Answer: B. 53

Answer 2

Answer:

B. 53

Step-by-step explanation: