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3 years ago

The sum of 5 consecutive integers is 265. What is the fifth number in this sequence?

2 answers:

7 0

Let x = the first consecutive number
Then x + 1 = the second consecutive number
Then x + 2 = the third consecutive number
Then x + 3 = the fourth consecutive number
Then x + 4 = the fifth consecutive number

The sum of the 5 consecutive integers is 265

x + x + 1 + x + 2 + x + 3 + x + 4 = 265

Combine like terms.

5x + 10 = 265

Solve for x.

5x + 10 = 265
5x = 255 <-- Subtract 10 from each side
x = 51 <-- Divide both sides by 5

So, the fifth consecutive integer is 51 + 4 or 55.

son4ous [18]3 years ago

3 0

Five consecutive integers are:

n, n+1, n+2, n+3, n+4 the sum of which is:

5n+10=265 subtract 10 from both sides

5n=255, divide both sides by 5

n=51

Our fifth number was n+4 so the fifth number is 55

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3 years ago

<img src="https://tex.z-dn.net/?f=Simplify%3A%20%5Cfrac%7B%205%C3%97%2825%29%5E%7Bn%2B1%7D%20-%2025%20%C3%97%20%285%29%5E%7B2n%7

Katen [24]

$\green{\large\underline{\sf{Solution-}}}$

Given expression is 

$\rm :\longmapsto\:\dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} }$

can be rewritten as

$\rm \: = \: \dfrac{5 \times { {(5}^{2} )}^{n + 1} - {5}^{2} \times {5}^{2n} }{5 \times {5}^{2n + 3} - {( {5}^{2} )}^{n + 1} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {( {x}^{m} )}^{n} \: = \: {x}^{mn}}}} \\$

And

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}} \\$

So, using this identity, we

$\rm \: = \: \dfrac{5 \times {5}^{2n + 2} - {5}^{2n + 2} }{{5}^{2n + 3 + 1} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 4} - {5}^{2n + 2} }$

can be further rewritten as

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 2 + 2} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2} - 1)}$

$\rm \: = \: \dfrac{4}{25 - 1}$

$\rm \: = \: \dfrac{4}{24}$

$\rm \: = \: \dfrac{1}{6}$

Hence, 

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} } = \frac{1}{6} }}$

4 0

3 years ago

Solve the system of linear equations.

sweet-ann [11.9K]

Answer:

dependent system
x = 2 -a
y = 1 +a
z = a

Step-by-step explanation:

Let's solve this by eliminating z, then we'll go from there.

Add 6 times the second equation to the first.

(3x -3y +6z) +6(x +2y -z) = (3) +6(4)

9x +9y = 27 . . . simplify

x + y = 3 . . . . . . divide by 9 [eq4]

Add 13 times the second equation to the third.

(5x -8y +13z) +13(x +2y -z) = (2) +13(4)

18x +18y = 54

x + y = 3 . . . . . . divide by 18 [eq5]

Equations [eq4] and [eq5] are identical. This tells us the system is dependent, and has an infinite number of solutions. We can find them in terms of z:

y = 3 -x . . . . solve eq5 for y

x +2(3 -x) -z = 4 . . . . substitute into the second equation

-x +6 -z = 4

x = 2 - z . . . . . . add x-4

y = 3 -(2 -z)

y = z +1

So far, we have written the solutions in terms of z. If we use the parameter "a", we can write the solutions as ...

x = 2 -a

y = 1 +a

z = a

_____

Check

First equation:

3(2-a) -3(a+1) +6a = 3

6 -3a -3a -3 +6a = 3 . . . true

Second equation:

(2-a) +2(a+1) -a = 4

2 -a +2a +2 -a = 4 . . . true

Third equation:

5(2-a) -8(a+1) +13a = 2

10 -5a -8a -8 +13a = 2 . . . true

Our solution checks algebraically.

6 0

3 years ago