Mathematics, 30.12.2019 07:31 shiannethorn

Pattern: 10, 17, 26,
what will the most likely number in this pattern be greater than 350?

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Answer from: 1hannacarson

362 is the 17th term of this pattern

Step-by-step explanation:

First differences of the sequence are ...

17-10 = 7, 26-17 = 9, 37 -26 = 11

And second differences are ...

9-7 = 2, 11-9 = 2

The constant second differences tell you the pattern is described by a second degree polynomial. That polynomial can be written as ...

an^2 +bn +c . . . for term number n

Filling in the first 3 sequence values, we get ...

a + b + c = 10

4a +2b +c = 17

9a +3b +c = 26

___

Any of several means can be used to solve these.

We can subtract the first equation from the other two to get ...

3a +b = 7 . . . . . . . . [eq4]

8a +2b = 16 . . . . . . [eq5]

Subtracting [eq4] from half of [eq5], we get ...

(4a +b) -(3a +b) = 8 -7

a = 1

Putting this value into [eq4] gives ...

3 +b = 7

b = 4

and using these values for a and b in the first equation gives ...

1 + 4 + c = 10

c = 5

So the pattern is described by ...

an = n^2 +4n +5

We want to find the term that is more than 350, so ...

350 < n^2 +4n +5

349 < n^2 +4n +4 . . . . subtract 1 to make a square on the right

349 < (n +2)^2

√349 ≈ 18.68 < n +2 . . . . take the positive square root

16.68 < n . . . . . . . . . . . subtract 2

The first term greater than 350 is the 17th term. It is 362.

In the process of "completing the square", we observe that the pattern rule can be written ...

an = (n+2)^2 +1

Answer from: Quest

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like, however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

$\bf e(\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-2}{)}\implies \implies \cfrac{-2}{0+2}\implies \cfrac{-2}{2}\implies -1 \\\\[-0.35em] ~\dotfill$

$\bf d(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{)}{)}\implies \cfrac{0+1}{0+2}\implies \cfrac{1}{2} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope~of~ef}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{1}\implies 1}}$

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

Answer from: Quest

cheater

step-by-step explanation:

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