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mrs_skeptik [129]
4 years ago
12

Evaluate the series: 2-10+50-250..., n=8

Mathematics
1 answer:
garik1379 [7]4 years ago
6 0

The sum of first 8 terms of series is -130208

<em><u>Solution:</u></em>

Given that,

2 - 10 + 50 - 250 , ....

n = 8

Find the common ratio between a term and its previous term

r = \frac{-10}{2} = -5\\\\r = \frac{-250}{50} = -5

Thus common ratio is same

This is a geometric series

<em><u>The sum of terms of geometric sequence is given as:</u></em>

S_n = \frac{a_1(1-r^n)}{1-r}

Where,

r is the common ratio

n is the nth term

a_1 is the first term of series

From series,

a_1 = 2\\\\r = -5\\\\n = 8

Therefore,

S_8 = \frac{2 \times (1 - (-5)^8)}{1-(-5)}\\\\S_8 = \frac{2 \times (1-390625)}{6}\\\\S_8 = -130208

Thus sum of first 8 terms of series is -130208

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What is the equation of a line that is perpendicular to −x+3y=9 and passes through the point (−3, 2) ?
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The equation of line that is perpendicular to the line -x+3y =9  and passes through the point \left({-3,2}\right)  is \boxed{{\mathbf{y=-3x-7}}} .

Further explanation:

It is given that the equation of line is -x+3y =9  and passes through point \left({-3,2}\right) .

Rewrite the given equation -x+3y =9  as follows:

\begin{aligned}-x+3y&=9\\3y&=9+x\\y&=\frac{9}{3}+\frac{1}{3}x\\y&=\frac{1}{3}x+3\\\end{aligned}

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It is given that both lines are perpendicular to each other so the product of slope must be equal to -1 .

{m_1}\cdot {m_2}=-1                                                          …… (1)

Substitute \frac{1}{3}  for {m_1}  in equation (1) to obtain the value of slope {m_2} .

\begin{aligned}\frac{1}{3}\cdot{m_2}&=-1\\{m_2}&=-3\\\end{aligned}

Therefore, the slope is -3 .

It is given that the line passes through point \left({-3,2}\right) .

The point-slope form of the equation of a line with slope m  passes through point \left({{x_1},{y_1}}\right) is represented as follows:

y-{y_1}=m\left({x-{x_1}}\right)                                      …… (2)

Substitute -3  for {x_1} , 2  for {y_1}  and -3  for m  in equation (2) to obtain the equation of line.

\begin{aligned}y-2&=-3\left({x-\left({-3}\right)}\right)\\y-2&=-3\left({x+3}\right)\\y&=-3x-9+2\\y&=-3x-7\\\end{aligned}

Therefore, the equation of line is y=-3x-7 .

Thus, the equation of line that is perpendicular to the line -x+3y=9  and passes through the point \left({-3,2}\right)  is \boxed{{\mathbf{y=-3x-7}}} .

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Answer Details:

Grade: Junior High School

Subject: Mathematics

Chapter: Coordinate Geometry

Keywords: Coordinate Geometry, linear equation, system of linear equations in two variables, variables, mathematics, equation of line, line, passes through point.

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