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2 years ago

Find the sum of the first 42 terms of the following series, to the nearest

Mathematics

1 answer:

balandron [24]2 years ago

5 0

Answer:

Step-by-step explanation:

Find the sum of the first 42 terms of the following series, to the nearest integer.

2,7,12

Solution

The sum is given by

SUM_n=n/2*(a_1+a_n)

a_n=a_1+(n-1)d

a_1=2, n=42, d=5

The 42nd term is therefore given by

a_42=2+(42-1)5=207

SUM_42=42/2*(2+207)=21*209=4389

The sum of the first 42 terms of the series, therefore, is 4389

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Can you exchange the coordinates of points ( x 1, y 1) and ( x 2, y 2) in the distance formula and still find the correct distan

notka56 [123]

Answer:

yes

Step-by-step explanation:

As an example

(x₁, y₁ ) = (1,2) and (x₂, y₂ ) = (5,3), then

d = $\sqrt{(5-1)^2+(3-2)^2}$ = $\sqrt{4^2+1^2}$ = $\sqrt{16+1}$ = $\sqrt{17}$

Now let (x₁, y₁ ) = (5,3) and (x₂, y₂ ) = (1,2), then

d = $\sqrt{(1-5)^2+(2-3)^2}$ = $\sqrt{(-4)^2+(-1)^2}$ = $\sqrt{16+1}$ = $\sqrt{17}$

3 0

3 years ago

Subtract 8 y^2 − 5 y + 7 from 2 y^2 + 7 y + 1 1 <br><br> The answer is: −6y ^2 +12y+4

lys-0071 [83]

Answer:

$\large\boxed{-6y^2+12y+4}$

Step-by-step explanation:

$(2y^2+7y+11)-(8y^2-5y+7)\\\\=2y^2+7y+11-8y^2-(-5y)-7\\\\=2y^2+7y+11-8y^2+5y-7\qquad\text{combine like terms}\\\\=(2y^2-8y^2)+(7y+5y)+(11-7)\\\\=-6y^2+12y+4$

4 0

3 years ago

Hello there ^^ I'd love it if you would help me :D

pochemuha

Do you have a question??

3 0

3 years ago

Read 2 more answers

Rational numbers that are not integers are?

Vladimir [108]

Any number that can be writen as a fraction example: 0.75, -15, 0, 9

6 0

3 years ago

If g(x) = 3x + 8, find g-1(4)

arsen [322]

Answer:

$\frac{-4}{3}$ =g-1(4)

Step-by-step explanation:

x=3y+8 ---> y= $\frac{x-8}{3}$

y= $\frac{(4)-8}{3}$

$\frac{-4}{3}$

4 0

2 years ago