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4 years ago

In a fibonacci sequence the fourth term is 30 and the fifth is 49 what is the first term

Mathematics

1 answer:

scoundrel [369]4 years ago

7 0

8 is the first.

Reasoning:

So we know what the Fibonacci sequence is. To get the next number, you add the number behind it, by the number behind that.

For example, these are the first 5 Fibonacci numbers.

1, 1, 2, 3, 5.

1 + 0 = 1.

So 1 is the second number.

1 + 1 is 2.

So 2 is the third number.

2 + 1 = 3.

So 3 is the fourth number.

3 + 2 = 5.

So 5 is the fourth number.

The problem:

If the fifth term is 49, and the fourth term is 30. We can subtract 49 by 30, to find the third number, and so on.

49 - 30 = 19.

So 19 is the third number.

30 - 19 = 11.

11 is the second.

19 - 11 = 8.

8 is the first.

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$\left (-\dfrac{4}{\sqrt{5}},\dfrac{8}{\sqrt{5}}\right )\text{ and }\left (\dfrac{4}{\sqrt{5}},-\dfrac{8}{\sqrt{5}}\right )$ .

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We need to find all the points on the curve x=4cos(t),y=4sin(t) that have the slope of 1/2.

$x=4cos (t)$

$\dfrac{dx}{dt}=-4sin (t)$

$y=4sin (t)$

$\dfrac{dy}{dt}=4cos (t)$

Now,

$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}$

$\dfrac{dy}{dx}=4cos (t)\times \dfrac{1}{-4sin (t)}$

$\dfrac{dy}{dx}=-\cot t$

So, slope of the curve is $-\cot t$ .

$-\cot t=\dfrac{1}{2}$

$-\tan t=2$

$\tan t=-2$ ...(1)

Using $\sec^2t=1+\tan^2t$ , we get

$\sec^2t=1+(-2)^2$

$\sec^2t=1+4$

$\sec t=\pm \sqrt{5}$

$\cos t=\pm \dfrac{1}{\sqrt{5}}$

Now,

$\sin^2t=1-cos^2t$

$\sin t=\pm \sqrt{1-\dfrac{1}{5}}$

$\sin t=\pm \sqrt{\dfrac{4}{5}}$

$\sin t=\pm \dfrac{2}{\sqrt{5}}$

It equation (1), tan(t) is negative. So, sin and cos have different signs.

If $\sin t= \dfrac{2}{\sqrt{5}}$ , then $\cos t=- \dfrac{1}{\sqrt{5}}$ .

$x=4cos (t)=-\dfrac{4}{\sqrt{5}}$

$y=4sin (t)=\dfrac{8}{\sqrt{5}}$

If $\sin t=- \dfrac{2}{\sqrt{5}}$ , then $\cos t= \dfrac{1}{\sqrt{5}}$ .

$x=4cos (t)=\dfrac{4}{\sqrt{5}}$

$y=4sin (t)=-\dfrac{8}{\sqrt{5}}$

Therefore, the two points are $\left (-\dfrac{4}{\sqrt{5}},\dfrac{8}{\sqrt{5}}\right )\text{ and }\left (\dfrac{4}{\sqrt{5}},-\dfrac{8}{\sqrt{5}}\right )$ .

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