Find the rule of the mapping: 1\;\;\;2\;\;\;3\;\;\;\;4\;\;\;\;5\;. . .\;x \;\;\;\;\;\;\;\;\;\;\;\; 7\;\;11\;\;15\;\;19\;\;23\;. . . .\;y

Correct Answer is : x⟶4x+3x 4 x + 3x⟶4x+3

Correct Answer is : x⟶4x+3x 4 x + 3x⟶4x+3

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BECE Mathematics 2010 Paper

Question : 24

Total: 40

1\;\;\;2\;\;\;3\;\;\;\;4\;\;\;\;5\;. . .\;x

\downarrow\;\;\downarrow\;\downarrow\;\downarrow\;\;\downarrow\;\;\;\;\;\;\downarrow

7\;\;11\;\;15\;\;19\;\;23\;. . . .\;y

Solution:

Note: If the mapping has equal interval (difference) for the

y

and

x

interval of 1 , you can use the general linear sequence formula to find the rule of the mapping.

U_n = U_1 + (n-1) d

In the mapping,

U_n = y

and

n = x

and

U_1

, the first value of

y

U_1 = 7

y = 7 + (x-1) d

The

d

is the interval between each consecutive

y

In this mapping,

d=11-7=15-11=19-15=23-19=4

We can substitute our

d

and simplify the expression for

y

y = 7 + (x-1) d

y = 7 + (x-1) 4

y = 7 + 4 \times x - 1 \times 4

y = 7 + 4 x - 4

y = 7 - 4 + 4 x

y = 3 + + 4 x

y = 4 x + 3

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