Asked by Aria

Arithmetic sequences

Find a8 and an for the sequence: -3, -7,-11

Answers

Answered by Scott
the common difference is -4

an = 1 - 4n
Answered by Aria
What is a8 ? Could it possibly be -31. Do i plug in 8 for the equation an=1-4n
Answered by MathMate
Scott has already given you the rule which is a<sub>n</sub>=1-4n.

To check the rule,
put n=1, a<sub>n</sub>=1-4=-3 (works)
put n=2, a<sub>n</sub>=1-4(2)=-7 (works)

To check a<sub>8</sub>, you can either use the given rule, or you can numerate the terms (8 is not too bad!)
-3, -7, -11, -15.....
Answered by Steve
well, let's try to take this step by step. Let's call the two lower vertices L and R. Suppose we have

<sub><sub></sub></sub>
π/3

L = (h,-k)
Then the line LC has slope
m<sub><sub>LC</sub></sub> = k/(1-h)
y+k = k/(1-h) (x-h)

Since LRC is equilateral, its angles are all π/3. So, the slope of LR is

m<sub><sub>LR</sub></sub> = tan(arctan(k/(1-h))-π/3)

and the equation of LR is

y+k = m<sub><sub>LR</sub></sub> (x-h)

Similarly, line RC has slope
m<sub><sub>RC</sub></sub> = tan(arctan(m<sub><sub>LR</sub></sub>)-π/3)

and its equation is (using point B)
y = tan(arctan(m<sub><sub>LR</sub></sub>)-π/3) (x-2)
Answered by Steve
well, let's try to take this step by step. Let's call the two lower vertices L and R. Suppose we have

L = (h,-k)
Then the line LC has slope
m<sub><sub>LC</sub></sub> = k/(1-h)
and its equation (using point A) is
y = k/(1-h) (x-1)

Since LRC is equilateral, its angles are all π/3. So, the slope of LR is

m<sub><sub>LR</sub></sub> = tan(arctan(k/(1-h))-π/3)

and the equation of LR is

y+k = m<sub><sub>LR</sub></sub> (x-h)

Similarly, line RC has slope
m<sub><sub>RC</sub></sub> = tan(arctan(m<sub><sub>LR</sub></sub>)-π/3)

and its equation is (using point B)
y = tan(arctan(m<sub><sub>LR</sub></sub>)-π/3) (x-2)
Answered by Steve
well, let's try to take this step by step. Let's call the two lower vertices L and R. Suppose we have

L = (h,-k)
Then the line LC has slope
m<sub><sub>LC</sub></sub> = k/(1-h)
and its equation (using point A) is
y = k/(1-h) (x-1)

Since LRC is equilateral, its angles are all π/3. So, the slope of LR is

m<sub><sub>LR</sub></sub> = tan(arctan(m<sub><sub>LC</sub></sub>)-π/3)

and the equation of LR is

y+k = m<sub><sub>LR</sub></sub> (x-h)

Similarly, line RC has slope
m<sub><sub>RC</sub></sub> = tan(arctan(m<sub><sub>LR</sub></sub>)-π/3)

and its equation is (using point B)
y = tan(arctan(m<sub><sub>LR</sub></sub>)-π/3) (x-2)

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