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nydimaria [60]
3 years ago
14

Find the sum of the first 20 terms of an arithmetic series if the first term is 4 and the common difference is 3.

Mathematics
1 answer:
Ratling [72]3 years ago
8 0

The sum of first 20 terms of Arithmetic Series is 650 if the first term is 4 and the common difference is 3.

Step-by-step explanation:

First Term  (a) = 4

Common difference  (d) = 3

The number of term (n) = 20

The sum of an Arithmetic series of (n) number of terms, with first term (a) and the common difference (d) is equal to

Sum =  (n/2) * ( 2 * a + (n -1) * d)

So putting the values of a,d, n

Sum = ( 20/2) * ( 2 * 4 + (20 -1) *3 )

Sum =  (10) * ( 8 + 19 * 3)

Sum =  10 * ( 8 + 57)

Sum = 10 * 65 = 650

Hence the sum of first 20 terms of Arithmetic Series is 650 if the first term is 4 and the common difference is 3.

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Un triángulo rectángulo tiene un área de 84 pies2 y una hipotenusa de 25 pies de largo. ¿Cuáles son las longitudes de sus otros
jolli1 [7]

Answer:

7 feet and 24 feet

Step-by-step explanation:

In the right triangle, the hypotenuse is 25 feet long. The area of this triangle is 84 square feet.

Let x feet and y feet be the lengths of triangle's legs.

By the Pythagorean theorem,

x^2+y^2=25^2\\ \\x^2+y^2=625

The area of the right triangle is half the product of its legs, thus

84=\dfrac{1}{2}xy\\ \\xy=168

Solve the system of two equations:

\left\{\begin{array}{l}x^2+y^2=625\\ \\xy=168\end{array}\right.

From the second equation:

x=\dfrac{168}{y}

Substitute it into the first equation:

\left(\dfrac{168}{y}\right)^2+y^2=625\\ \\168^2+y^4=625y^2\\ \\y^4-625y^2+168^2=0\\ \\D=(-625)^2-4\cdot 168^2=(625-2\cdot 168)(625+2\cdot 168)=289\cdot 961=17^2\cdot 31\\ \\\sqrt{D}=17\cdot 31=527\\ \\y^2_{1,2}=\dfrac{-(-625)\pm 527}{2}=49,\ 576\\ \\y_1=7,\ y_2=-7,\ y_3=24,\ y_4=-24

The length of the leg cannot be negative, so

y_1=7\Rightarrow x_1=\dfrac{168}{7}=24\\ \\y_2=24\Rightarrow x_2=\dfrac{168}{24}=7

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Step-by-step explanation:

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