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3 years ago

The sum of three consecutive even integers is 90. Find the Integers.

Mathematics

2 answers:

Shtirlitz [24]3 years ago

7 0

x + x + 1 + x + 2 = 90 ....

3x+3= 90

meaning, 3x= 87

x=27

The integers are 29, 30 and 31.

I hope I could help! :)

viva [34]3 years ago

4 0

Answer:28,30,32

I had answer this to soneone else already but here it is

Step-by-step explanation:

The sum of three consecutive even integers is 90. Find the Integers.

N= first even number we need to find

N+2= second even number

N+4= third even number

90= is the sum of all 3 even integers

N+n+2+n+4=90

Add all n together

3n+2+4=90

Add 2 plus 4

3n+6=90

Subtract 6 bith sides

3n=90-6

Subtract 90 from 6

3n= 84

Divide 3 both sides

3n/3=84/3

Divide 84 by 3 to solve for n

N= 28

So furst even number is 28. Now to find second and third number substitute 28 into the bext two equations .

N+2=? For second numb.

28+2=30 second number

N+4=? For third number

28+4=32 for third number

To check add all 3 answers

28+30+32=90

58+32=90

90=90 it checks

So three even numbers are 28,30,32

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Find the number of terms, n, in the arithmetic series whose first term is 13, the common difference is 7, and the sum is 2613.

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Step-by-step explanation:

Recall that the sum of an arithmetic series is given by:

$\displaystyle S = \frac{n}{2}\left(a + x_n\right)$

Where n is the number of terms, a is the first term, and x_n is the last term.

We know that the initial term a is 13, the common difference is 7, and the total sum is 2613. Since we want to find the number of terms, we want to find n.

First, find the last term. Recall that the direct formula for an arithmetic sequence is given by:

$x_n=a+d(n-1)$

Since the initial term is 13 and the common difference is 7:

$x_n=13+7(n-1)$

Substitute:

$\displaystyle S = \frac{n}{2}\left(a + (13+7(n-1)\right)$

We are given that the initial term is 13 and the sum is 2613. Substitute:

$\displaystyle (2613)=\frac{n}{2}((13)+(13+7(n-1)))$

Solve for n. Multiply both sides by two and combine like terms:

$5226 = n(26+7(n-1))$

Distribute:

$5226 = n (26+7n-7)$

Simplify:

$5226 = 7n^2+19n$

Isolate the equation:

$7n^2+19n-5226=0$

We can use the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 7, b = 19, and c = -5226. Substitute:

$\displaystyle x =\frac{-(19)\pm\sqrt{(19)^2-4(7)(-5226)}}{2(7)}$

Evaluate:

$\displaystyle x = \frac{-19\pm\sqrt{146689}}{14} = \frac{-19\pm 383}{14}$

Evaluate for each case:

$\displaystyle x _ 1 = \frac{-19+383}{14} = 26\text{ or } x _ 2 = \frac{-19-383}{14}=-\frac{201}{7}$

We can ignore the second solution since it is negative and non-natural.

Therefore, there are 26 terms in the arithmetic series.

Our answer is A.

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