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Licemer1 [7]
4 years ago
9

3 + x, 9 + 3x, 13 + 4x, ... is an arithmetic sequence for some real number x.

Mathematics
1 answer:
omeli [17]4 years ago
5 0

Answer:

x = -2

Step-by-step explanation:

An arithmetic sequence is one where the difference between any two consecutive terms in the sequence is constant.

For any arithmetic sequence the general formula is as follow

a_{n} = a_{1} + (n - 1)d

where

a_{n} is the nth arithmetic value

a_{1} is the initial arithmetic value

n is the nth term

d is the difference is consecutive arithmetic values

To find the value of x we form a number of simultaneous equations

a_{1} = 3 + x\\ a_{2} = (3 + x) + 1 d\\ a_{3} = (3 + x) + 2 d

from the given information we also know that

a_{2} = 9 + 3 x\\  a_{3} = 13 + 4x

Solving these equations simultaneous we find the value of x to be

x = -2

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FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

cos \beta = -\frac{1}{2}  \sqrt{3}

tan \beta= \frac{1}{3}  \sqrt{3}

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sin (α + β) = sin α cos β + cos α sin β
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sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )
sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )
cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )
cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )

Find tan (α - β)
tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha  tan \beta }
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3}   }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}

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tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3}   }{1+(\frac{5\sqrt{3}}{24})}
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }

Simplify the numerator
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }
tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }

Simplify the fraction
tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})
tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}

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