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3 years ago

Find the nth term of the sequence 7,25,51,85,127

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

4 0

Let a (n) denote the n-th term of the given sequence.

Check the forward differences, and denote the n-th difference by b (n). That is,

b (n) = a (n + 1) - a (n)

These so-called first differences are

b (1) = a (2) - a (1) = 25 - 7 = 18

b (2) = a (3) - a (2) = 51 - 25 = 26

b (3) = a (4) - a (3) = 85 - 51 = 34

b (4) = a (5) - a (4) = 127 - 85 = 42

Now consider this sequence of differences,

18, 26, 34, 42, …

and notice that the difference between consecutive terms in this sequence b is 8:

26 - 18 = 8

34 - 26 = 8

42 - 34 = 8

and so on. This means b is an arithmetic sequence, and in particular follows the rule

b (n) = 18 + 8 (n - 1) = 8n + 10

for n ≥ 1.

So we have

a (n + 1) - a (n) = 8n + 10

or, replacing n + 1 with n,

a (n) = a (n - 1) + 8 (n - 1) + 10

a (n) = a (n - 1) + 8n + 2

We can solve for a (n) by iteratively substituting:

a (n) = [a (n - 2) + 8 (n - 1) + 2] + 8n + 2

a (n) = a (n - 2) + 8 (n + (n - 1)) + 2×2

a (n) = [a (n - 3) + 8 (n - 2) + 2] + 8 (n + (n - 1)) + 2×2

a (n) = a (n - 3) + 8 (n + (n - 1) + (n - 2)) + 3×2

and so on. The pattern should be clear; we end up with

a (n) = a (1) + 8 (n + (n - 1) + … + 3 + 2) + (n - 1)×2

The middle group is the sum,

$\displaystyle 8\sum_{k=2}^nk=8\sum_{k=1}^nk-8=\frac{8n(n+1)}2-8=4n^2+4n-8$

so that

a (n) = a (1) + (4n ² + 4n - 8) + 2 (n - 1)

a (n) = 4n ² + 6n - 3

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3 years ago

Find the equation of a line passing through points (-7, -10) , (-5, -20)

LuckyWell [14K]

You want to find the equation for a line that passes through the two points:

(-7,-10) and (-5,-20).

First of all, remember what the equation of a line is:

y = mx+b

here, m is the slope, b is the y-intercept

First, let's find what m is, the slope of the line...

The slope of a line is a measure of how fast the line "goes up" or "goes down". A large slope means the line goes up or down really fast (a very steep line). Small slopes means the line isn't very steep. A slope of zero means the line has no steepness at all; it is perfectly horizontal.

For lines like these, the slope is always defined as "the change in y over the change in x" or, in equation form:

So what we need now are the two points you gave that the line passes through.

Consider (-7,-10) as point #1, so the x and y numbers given will be called x1 and y1. Or, x1=-7 and y1=-10.

Consider (-5,-20), point #2, so the x and y numbers here will be called x2 and y2. Or, x2=-5 and y2=-20.

Now, just plug the numbers into the formula for m above, like this:

m= (-20 - -10)/(-5 - -7)

m= -10/2

m=-5

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

y=-5x+b

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(-7,-10). When x of the line is -7, y of the line must be -10.

(-5,-20). When x of the line is -5, y of the line must be -20.

Because line passes through each one of these two points, right?

Now, look at our line's equation so far: y=-5x+b. b is what we want, the -5 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specifically passes through the two points (-7,-10) and (-5,-20).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!.

You can use either (x,y) point you want.The answer will be the same:

(-7,-10). y=mx+b or -10=-5 × -7+b, or solving for b: b=-10-(-5)(-7). b=-45.

(-5,-20). y=mx+b or -20=-5 × -5+b, or solving for b: b=-20-(-5)(-5). b=-45.

See! In both cases we got the same value for b. And this completes our problem.

The equation of the line that passes through the points (-7,-10) and (-5,-20) is y=-5x-45.

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3 years ago

Draw a line having a slope of 1/3 and y-intercept of -1

Lilit [14]

Equation for slope=1/3 and y-intercept=-1 is:
y = mx + b
where m is slope and b is y-intercept.
So, equation becomes
y = -1x + 1/3
Now put different values of x in the equation to get corresponding value of y.
x     y
0    1/3
1 -2/3
2 -5/3
3 -8/3
-1    4/3
-2    7/3
-3    10/3

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3 years ago

How do you do estimate of percent problems?

vovangra [49]

How ever many times it goes into 100

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3 years ago

La potencia que se obtiene de elevar a un mismo exponente un numero racional y su opuesto es la misma verdadero o falso?

malfutka [58]

Answer:

Falso.

Step-by-step explanation:

Sea $d = \frac{a}{b}$ un número racional, donde $a, b \in \mathbb{R}$ y $b \neq 0$ , su opuesto es un número real $c = -\left(\frac{a}{b} \right)$ . En el caso de elevarse a un exponente dado, hay que comprobar cinco casos:

(a) El exponente es cero.

(b) El exponente es un negativo impar.

(c) El exponente es un negativo par.

(d) El exponente es un positivo impar.

(e) El exponente es un positivo par.

(a) El exponente es cero:

Toda potencia elevada a la cero es igual a uno. En consecuencia, $c = d = 1$ . La proposición es verdadera.

(b) El exponente es un negativo impar:

Considérese las siguientes expresiones:

$d' = d^{-n}$ y $c' = c^{-n}$

Al aplicar las definiciones anteriores y las operaciones del Álgebra de los números reales tenemos el siguiente desarrollo:

$d' = \left(\frac{a}{b} \right)^{-n}$ y $c' = \left[-\left(\frac{a}{b} \right)\right]^{-n}$

$d' = \left(\frac{a}{b} \right)^{(-1)\cdot n}$ y $c' = \left[(-1)\cdot \left(\frac{a}{b} \right)\right]^{(-1)\cdot n}$

$d' = \left[\left(\frac{a}{b} \right)^{-1}\right]^{n}$ y $c' = \left[(-1)^{-1}\cdot \left(\frac{a}{b} \right)^{-1}\right]^{n}$

$d' = \left(\frac{b}{a} \right)^{n}$ y $c = (-1)^{n}\cdot \left(\frac{b}{a} \right)^{n}$

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = \left[(-1)\cdot \left(\frac{b}{a} \right)\right]^{n}$

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = \left[-\left(\frac{b}{a} \right)\right]^{n}$

Si $n$ es impar, entonces:

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = - \left(\frac{b}{a} \right)^{n}$

Puesto que $d' \neq c'$ , la proposición es falsa.

(c) El exponente es un negativo par.

Si $n$ es par, entonces:

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = \left(\frac{b}{a} \right)^{n}$

Puesto que $d' = c'$ , la proposición es verdadera.

(d) El exponente es un positivo impar.

Considérese las siguientes expresiones:

$d' = d^{n}$ y $c' = c^{n}$

$d' = \left(\frac{a}{b}\right)^{n}$ y $c' = \left[-\left(\frac{a}{b} \right)\right]^{n}$

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = \left[(-1)\cdot \left(\frac{a}{b} \right)\right]^{n}$

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = (-1)^{n}\cdot \left(\frac{a}{b} \right)^{n}$

Si $n$ es impar, entonces:

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = - \left(\frac{a}{b} \right)^{n}$

(e) El exponente es un positivo par.

Considérese las siguientes expresiones:

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = \left(\frac{a}{b} \right)^{n}$

Si $n$ es par, entonces $d' = c'$ y la proposición es verdadera.

Por tanto, se concluye que es falso que toda potencia que se obtiene de elevar a un mismo exponente un número racional y su opuesto es la misma.

3 0

3 years ago

Find the nth term of the sequence 7,25,51,85,127​

Find the nth term of the sequence 7,25,51,85,127