Question

a sequence has a first number of 3 and a half common difference of 4 what is the recursive rule of nth term​

Answer 1

Answer:

$f(n) = f(n-1) + 8$ for $n > 1$

Step-by-step explanation:

Given

$f(1) = 3$ -- First Term

$\frac{1}{2}d = 4$ --- half common difference

Required

Find the recursive rule

First, we calculate the common difference

$\frac{1}{2}d = 4$

Multiply through by 2

$2 * \frac{1}{2}d = 2 * 4$

$d = 8$

The second term of the sequence is:

$f(2) = 3 + 8 = 11$

The third term is:

$f(3) = 11 + 8 = 20$

So, we have:

$f(1) = 3$

$f(2) = 3 + 8$

Substitute f(1) for 3

$f(2) = f(1) + 8$

Express 1 as 2 - 1

$f(2) = f(2-1) + 8$

Substitute n for 2

$f(n) = f(n-1) + 8$

Similarly:

$f(3) = 11 + 8$

Substitute f(2) for 11

$f(3) = f(2) + 8$

Express 2 as 3 - 1

$f(3) = f(3-1) + 8$

Substitute n for 3

$f(n) = f(n-1) + 8$

Hence, the recursive is:

$f(n) = f(n-1) + 8$ for $n > 1$