What is the seventh term of the sequence whose nth term is (n + 1)(n − 2)?
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Answer:
(0.5,3.5)
Step-by-step explanation:
First, we can draw the image, as shown. The diagonals in the rectangle are the following lines:
from (-2,9) to (3,-2)
from (-5, 6) to (6,1)
To find where they intersect, we can start by making an equation for the lines. For an equation y=mx+b, m represents the slope and b represents the y intercept, or when x=0
For the first line, from (-2,9) to (3,-2), we can calculate the slope by calculating the change in y/change in x = (y₂-y₁)/(x₂-x₁). If (3,-2) is (x₂,y₂) and (-2,9) is (x₁,y₁), our slope is
(-2-9)/(3-(-2)) = -11/5
Therefore, our equation is
y= (-11/5)x + b
To solve for b, we can plug a point in, like (3,-2). Therefore,
-2=(-11/5)*3+b
-2=-33/5+b
-10/5=-33/5+b
add 33/5 to both sides to isolate b
23/5=b
Our equation for one diagonal is therefore y=(-11/5)x+23/5
For the second line, from (-5, 6) to (6,1), if (6,1) is (x₁,y₁) and (-5,6) is (x₂,y₂), the slope is (1-6)/(6-(-5)) = -5/11 . Plugging (6,1) into the equation y=(-5/11)x+b, we have
1=(-5/11)*6+b
11/11 = -30/11 + b
add 30/11 to both sides to isolate b
41/11 = b
our equation is
y = (-5/11) x + 41/11
Our two equations are thus
y = (-5/11) x + 41/11
y=(-11/5)x+23/5
To find where they intersect, we can set them equal to each other
(-11/5)x+23/5 = y = (-5/11) x + 41/11
(-11/5)x + 23/5 = (-5/11)x + 41/11
subtract 23/5 from both sides as well as add 5/11 to both sides to make one side have only x values and their coefficients
(-11/5)x + (5/11)x = 41/11-23/5
11*5 = 55, so 55 is one value we can use to make the denominators equal.
(-11*11/5*11)x+(5*5/11*5)x=(41*5/11*5)-(23*11/5*11)
(-121/55)x+(25/55)x = (205/55) - (253/55)
(-96/55)x = (-48/55)
multiply both sides by 55 to remove the denominators
-96x=-48
divide both sides by -96 to isolate x
x=-48/-96=0.5
plug x=0.5 into a diagonal to see the y value of the intersection
(-11/5)x + 23/5 = y = (-11/5)* 0.5 + 23/5 = 3.5
