+0

# flow chart

0
117
3
+846

Inputs are 1; 3; 5 ;7 (X values)

Outputs are 5; 10; 15; 20 (Y values)

I did the following:

constant difference is according to the outputs, which is 5.

the formule used is y=bx+c, b equal to the constant difference

so we have y=5x+c.

So if I take the first term for example, to calculate "c"

y=5(x)+c

5=5(1)+c

c=0

however, using c=0, does not calculate correctly for the other terms. also, "c" calculates differently for each term. Please help, I really do appreciate.

May 23, 2021

#1
+26213
+3

Inputs are 1; 3; 5 ;7 (X values)

Outputs are 5; 10; 15; 20 (Y values)

$$\begin{array}{|r|r|r|r|} \hline x & \dfrac{x+1}{2} & \left(\dfrac{x+1}{2}\right) \times 5 \\ \hline 1 & \dfrac{1+1}{2}= 1 & 1*5 = 5 \\ \hline 3 & \dfrac{3+1}{2}= 2 & 2*5 = 10\\ \hline 5 & \dfrac{5+1}{2}= 3 & 3*5 = 15\\ \hline 7 & \dfrac{7+1}{2}= 4 & 4*5 = 20 \\ \hline \end{array}$$

The rule is $$y= \left(\dfrac{x+1}{2}\right)\times 5$$

May 23, 2021
#2
+846
+1

Hi heureka,

Thank you for the response, however, could you kindly explain how you got to the rule in the first place...how did you know it had to be $${(x+1)\over2}*5$$

juriemagic  May 23, 2021
#3
+26213
+2

how you got to the rule in the first place:

Arithmetic progression: $$1,~ 3,~ 5,~ 7,~\ldots$$
$$\begin{array}{|rcll|} \hline a_n &=& a_1+(n-1)*d \quad | \quad a_1 = 1,~ d=2 \\ a_n &=& 1+(n-1)*2 \\ a_n &=& 1+2n -2 \\ a_n &=& 2n-1 \\ 2n &=& a_n + 1 \\ n &=& \dfrac{a_n+1}{2} \quad \text{or} \quad n = \dfrac{x+1}{2}\\ \hline \end{array}$$

$$\dots~ y = 5n$$

heureka  May 23, 2021
edited by heureka  May 23, 2021
edited by heureka  May 23, 2021