Question

The pattern of numbers below is an arithmetic sequence: 14, 24, 34, 44, 54, ... Which statement describes the recursive function used to generate the sequence? A. The common difference is 1, so the function is f(n + 1) = f(n) + 1 where f(1) = 14. B. The common difference is 4, so the function is f(n + 1) = f(n) + 4 where f(1) = 10. C. The common difference is 10, so the function is f(n + 1) = f(n) + 10 where f(1) = 14. D. The common difference is 14, so the function is f(n + 1) = f(n) + 14 where f(1) = 10.

Answer 1

Answer:

Option C is correct

The common difference is 10,

So,  the function is f(n + 1) = f(n) + 10 , where f(1) = 14

Step-by-step explanation:

The recursive function for the arithmetic sequence is given by:

$f(n+1) = f(n)+d$           .....[1]

where, d is the common difference of the two consecutive terms.

Given the arithmetic sequence :

14, 24, 34, 44, 54, .......

First term f(1) = 14

Common difference(d) = 10

Since,

24 -14 = 10

34-24 = 10

44-34 = 10 and so on....

Substitute d = 4 in [1], we have;

$f(n+1) = f(n) + 10$

Therefore, the recursive function used to generate the sequence is,

$f(n+1) = f(n) + 10$ and f(1) = 10

Answer 2

The common difference is 10, so the function is t(n + 1) = t(n) + 10. Option C is correct.

Given the sequence of numbers  14, 24, 34, 44, 54, ...

This can also be written as (4+10), (14+10) + (24+10), (34+10), (44+10)...

Since 10 is comon to all the terms, the recursive formula will be:

t(n+1) = t(n) + 10

t(n) is the nth term of the remaining sequence 4, 14, 24, 34...

t(n) = a+(n-1)d

t(n) = 4+(n-1)(10)

t(n) = 4 + 10n - 10

t(n) = 10n - 6

Hence the equivalent recursive function will be t(n + 1)= t(n) + 10 with a common difference of 10

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