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damaskus [11]
3 years ago
15

Identify the following series as arithmetic, geometric, both, or neither.

Mathematics
2 answers:
Strike441 [17]3 years ago
6 0

Answer:

It is a Geometric series.

Step-by-step explanation:

It has a common ratio equal to a:

3a^2 / 3a = a, 3a^3 / 3a^2 = a  and so on..

It is a Geometric series.

pav-90 [236]3 years ago
6 0

Answer with explanation:

 The given series is:

      3 a + 3 a^2 + 3 a^3 + . . . + 3 a^n

If you will try out to find out the ratio of

    \rightarrow\frac{\text{Second term}}{\text{First term}} {\text{or}}\frac{\text{Third term}}{\text{Second term}}  {\text{or}} \frac{\text{Fourth term}}{\text{Third term}} ........\\\\ \rightarrow\frac{3a^2}{3a}=\frac{3a^3}{3a^2}=\frac{3a^4}{3a^3}=......=a

The Ratio of Succeeding term to it's preceding term in the given sequence is constant equal to a.

So, if any Sequence follows this kind of rule or pattern we call it Geometric Progression.

Option B:→ Geometric

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Define the mathematical base <br>And give an example
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HOPE IT HELPED

6 0
3 years ago
A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea
Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let <em>x</em> be the width and <em>y </em>the length of the rectangular field.

Let <em>C </em>the total cost of the rectangular field.

The side made of heavy duty material of length of <em>x </em>costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are <em>x, y, y</em>.  Thus

C=4x+4y+4y+16x\\C=20x+8y

We know that the total area of rectangular field should be 2250 square yards,

x\cdot y=2250

We can say that y=\frac{2250}{x}

Substituting into the total cost of the rectangular field, we get

C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}

We have to figure out where the function is increasing and decreasing. Differentiating,

\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}

Next, we find the critical points of the derivative

20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30

Because the length is always positive the only point we take is x=30. We thus test the intervals (0, 30) and (30, \infty)

C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0

we see that total cost function is decreasing on (0, 30) and increasing on (30, \infty). Therefore, the minimum is attained at x=30, so the minimal cost is

C(30)=20(30)+\frac{18000}{30}\\C(30)=1200

The least cost of fencing for the rancher is $1200

Here’s the diagram:

3 0
3 years ago
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