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3 years ago

What is the 48th term of the arithmetic sequence with this explicit formula?

Mathematics

1 answer:

mariarad [96]3 years ago

4 0

Option B

48th term of arithmetic sequence is -152

<h3><u>Solution:</u></h3>

Given that a arithmetic sequence has explict formula,

$a_n = -11 + (n - 1)(-3)$

To find: 48th term of the arithmetic sequence

The nth term of arithmetic sequence can be found by substituting the required value of n in the explict formula

So 48th term of the arithmetic sequence can be found by substituting n = 48 in explict formula

$a_n = -11 + (n - 1)(-3)$

Plug in n = 48

$a_{48} = -11 + (48 - 1)(-3)$

On solving we get,

$a_{48} = -11 + (47)(-3)\\\\a_{48} = -11 - 141\\\\a_{48} = - 152$

Thus 48th term of arithmetic sequence is -152

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A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 ft/s from a point 500

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Answer:

The rate at which both of them are moving apart is 4.9761 ft/sec.

Step-by-step explanation:

Given:

Rate at which the woman is walking, $\frac{d(w)}{dt}$ = 3 ft/sec

Rate at which the man is walking, $\frac{d(m)}{dt}$ = 2 ft/sec

Collective rate of both, $\frac{d(m+w)}{dt}$ = 5 ft/sec

Woman starts walking after 5 mins so we have to consider total time traveled by man as (5+15) min = 20 min

Now,

Distance traveled by man and woman are $m$ and $w$ ft respectively.

⇒ $m=2\ ft/sec=2\times \frac{60}{min} \times 20\ min =2400\ ft$

⇒ $w=3\ ft/sec = 3\times \frac{60}{min} \times 15\ min =2700\ ft$

As we see in the diagram (attachment) that it forms a right angled triangle and we have to calculate $\frac{dh}{dt}$ .

Lets calculate h.

Applying Pythagoras formula.

⇒ $h^2=(m+w)^2+500^2$

⇒ $h=\sqrt{(2400+2700)^2+500^2} = 5124.45$

Now differentiating the Pythagoras formula we can calculate the rate at which both of them are moving apart.

Differentiating with respect to time.

⇒ $h^2=(m+w)^2+500^2$

⇒ $2h\frac{d(h)}{dt}=2(m+w)\frac{d(m+w)}{dt} + \frac{d(500)}{dt}$

⇒ $\frac{d(h)}{dt} =\frac{2(m+w)\frac{d(m+w)}{dt} }{2h}$ ...as $\frac{d(500)}{dt}= 0$

⇒ Plugging the values.

⇒ $\frac{d(h)}{dt} =\frac{2(2400+2700)(5)}{2\times 5124.45}$ ...as $\frac{d(m+w)}{dt} = 5$ ft/sec

⇒ $\frac{d(h)}{dt} =4.9761$ ft/sec

So the rate from which man and woman moving apart is 4.9761 ft/sec.

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