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3 years ago

Find the sum of the first 25 terms of the arithmetic sequnce 17,22,27,32.37

Mathematics

2 answers:

crimeas [40]3 years ago

7 0

Answer: The sum of the first 25 terms of this arithmetic sequence is 1925.

Explanation: This sequence, beginning at 17, increases by 5 per term. By simply adding 5 to the preceding term in the sequence, you can find the next term. The first term is 17, and the last term is 137. This means the list of terms is 17, 22, 27, 32, 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97, 102, 107, 112, 117, 122, 127, 132, and 137. The sum of these terms is 1925.

ANTONII [103]3 years ago

6 0

Answer:

Step-by-step explanation:

This sequence is adding 5 so

17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97 102 107 112 117 122 127 132 137 = 1925

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A small radio transmitter broadcasts in a 30 mile radius. If you drive along a straight line from a city 33 miles north of the t

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34.26miles from 49.58miles of the drive (69.11% of the drive).

Step-by-step explanation:

A problem of Analytic Geometry. This question can be solved using the resulting values from equaling the equations of the line (for the drive) and circle (with radius equals 30miles for radio transmitter broadcasting), and calculating the total distances from coincident points between circle and line, and total drive.

A graph is attached showing part of the circle and line with coincident points.

<h3>Line's Equation</h3>

Assuming transmitter is located (0, 0). From a city 33 miles north of the transmitter (0, 33) to a second city 37 miles east of the transmitter (37, 0).

Then, the equation for the line is (taking the two points from the start to the end):

$\\ y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}*(x-x_{1})$

$\\ y-33=\frac{0-33}{37-0}*(x-0)}$

$\\ y-33=\frac{-33}{37}*x}$

$\\ y=\frac{-33}{37}*x} + 33$ [1]

<h3>Circle's Equation</h3>

The equation for the circle whose center (0, 0) is:

$\\ (x-0)^2 + (y-0)^2 = 30^2$

$\\ x^2 + y^2 = 30^2$

$\\ y^2 = 30^2 - x^2$

$\\ y = \frac{+}{-}\sqrt{30^2- x^2}$ [2]

Equaling [1] and [2], to determine the points where starting to receive the radio transmitter signals to the point where finishing those signals:

$\\ \frac{-33}{37}*x + 33 = \sqrt{30^2- x^2}$

Solving this equation for <em>x</em>, we have two solutions for it (from <em>WolframAlpha</em>):

$\\ x_{1} = 40293/2458 - (111 \sqrt(80151))/2458 \approx 3.60775$

$\\ x_{2} = 40293/2458 + (111 \sqrt(80151))/2458 \approx 29.1774$

Then, using [1], the corresponding values for <em>y</em> are:

$\\ y_{1}=\frac{-33}{37}*(3.60775) \approx 29.78$

$\\ y_{2}=\frac{-33}{37}*(29.1774) \approx 6.97$

So,

$\\ (x_{1}=3.61, y_{1}=29.78)$

$\\ (x_{2}=29.18, y_{2}=6.97)$

Well, the distance of the drive is:

$\\ d = \sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}$

$\\ d = \sqrt{(29.18-3.61)^2 + (6.97-29.78)^2}$

$\\ d \approx 34.2654$ miles.

The total distance traveled is:

$\\ d = \sqrt{(37-0)^2 + (0-33)^2}$

$\\ d \approx 49.5782$ miles

Thus, during the drive, the signal of the radio transmitter was picked up for 34.26miles from the total of the 49.58 traveled, that is, a fraction equivalent to $\\ \frac{34.2654}{49.5782} \approx 0.6911$ or 69.11% of the drive.

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