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3 years ago

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Let g be a geometric sequence of integers with first term 1, and let G be the series associated with g. Let the natural number n

be such that gn=78125 and Gn=97656. What is the value of n? GEOMETRIC SEQUENCES

1 answer:

avanturin [10]3 years ago

8 0

Answer:

n = 8

Step-by-step explanation:

If the common ratio is r, then the term gn is ...

gn = 1·r^(n-1)

and the sum Gn is ...

Gn = 1·(r^n -1)/(r -1)

Using the given values for gn and Gn, we can write ...

r^n = r·gn

so ...

Gn = 97656 = (78125r -1)/(r -1)

Multiplying by r-1 gives

97656r -97656 = 78125r -1

19531r = 97655 . . . . . add 97656-78125r

r = 5 . . . . . . . . . . . . . . divide by 19531

___

Now we can find the value of n. Let's use gn.

78125 = 5^(n-1) = 5^7

Matching exponents, we find ...

n -1 = 7

n = 8 . . . . . . add 1

The value of n is 8.

_____

<em>Comment on power of 5</em>

You can use logarithms to find what power of 5 gets you 78125. Taking the log of the above equation, you get

log(78125) = (n-1)log(5)

log(78125)/log(5) = n-1

7 = n -1 . . . . . . evaluate the ratio of the logs

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Answer:

The area function is

$A=\frac{135}{2}x-\frac{9}{2}x^2$ .

The domain and range of A is $(0,15m)$ and $(0, 253.125 m^2]$ .

Step-by-step explanation:

The given length of fencing is $90 m$ .

Let the length and width of each pen be $x$ and $y$ respectively as shown in the figure.

As there are 3 pens, so, the total area,

$A= 3 xy \;\cdots (i)$

From the figure the total length of fencing is $6x+4y$ .

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From the given value:

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$\Rightarrow y=\frac {45}{2}-\frac{3x}{2}$

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$A=3x\left(\frac {45}{2}-\frac{3x}{2}\right)$

$\Rightarrow A=\frac{135}{2}x-\frac{9}{2}x^2\;\cdots (iii)$

This is the required area function in the terms of variable $x$ .

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For the range of area function:

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Now, differentiate the area function with respect to $x$ .

$\frac {dA}{dx}=\frac{135}{2}-9x$

Equate $\frac {dA}{dx}$ to zero to get the extremum point.

$\frac {dA}{dx}=0$

$\Rightarrow \frac{135}{2}-9x=0$

$\Rightarrow x=\frac{15}{2}$

Check this point by double differentiation

$\frac {d^2A}{dx^2}=-9$

As, $\frac {d^2A}{dx^2}$ , so, point $x=\frac{15}{2}$ is corresponding to maxima.

Put this value back to equation (iii) to get the maximum value of area function. We have

$A=\frac{135}{2}\times \frac {15}{2}-\frac{9}{2}\times \left(\frac {15}{2}\right)^2$

$\Rightarrow A=253.125 m^2$

Hence, the range of area function is $(0, 253.125 m^2]$ .

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