Arithmetic sequences

Question

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

Answer 1

the common difference is -4

an = 1 - 4n

Answer 2

What is a8 ? Could it possibly be -31. Do i plug in 8 for the equation an=1-4n

Answer 3

Scott has already given you the rule which is a_n=1-4n.

To check the rule,
put n=1, a_n=1-4=-3 (works)
put n=2, a_n=1-4(2)=-7 (works)

To check a₈, you can either use the given rule, or you can numerate the terms (8 is not too bad!)
-3, -7, -11, -15.....

Answer 4

well, let's try to take this step by step. Let's call the two lower vertices L and R. Suppose we have


π/3

L = (h,-k)
Then the line LC has slope
m_LC = 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_LR = tan(arctan(k/(1-h))-π/3)

and the equation of LR is

y+k = m_LR (x-h)

Similarly, line RC has slope
m_RC = tan(arctan(m_LR)-π/3)

and its equation is (using point B)
y = tan(arctan(m_LR)-π/3) (x-2)

Answer 5

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_LC = 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_LR = tan(arctan(k/(1-h))-π/3)

and the equation of LR is

y+k = m_LR (x-h)

Similarly, line RC has slope
m_RC = tan(arctan(m_LR)-π/3)

and its equation is (using point B)
y = tan(arctan(m_LR)-π/3) (x-2)

Answer 6

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_LC = 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_LR = tan(arctan(m_LC)-π/3)

and the equation of LR is

y+k = m_LR (x-h)

Similarly, line RC has slope
m_RC = tan(arctan(m_LR)-π/3)

and its equation is (using point B)
y = tan(arctan(m_LR)-π/3) (x-2)