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Alexandra [31]
2 years ago
11

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

Mathematics
1 answer:
ELEN [110]2 years ago
8 0

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

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Answer:

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Step-by-step explanation:

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hope this helps

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