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Lunna [17]
3 years ago
13

The first term of a geometric sequence is –2 and the common ratio is -1/4 . What are the next three terms of the sequence

Mathematics
2 answers:
rodikova [14]3 years ago
5 0

well -2(-1/4) is .5

so -2, .5,(multiply each by -1/4)

-2, 0.5, -0.125, 0.03125

Mumz [18]3 years ago
3 0

Answer:

\text{The next three terms of G.P is }\frac{1}{2}, \frac{-1}{8}, \frac{1}{32}

Step-by-step explanation:

Given that the first term of a geometric sequence is -2 and the common ratio is -1/4.

we have to find the next three terms of the sequence.

The geometric sequence can be written as

a, ar, ar^2, ar^3, ar^4,...

where a and r are the first term and common ratio.

a=-2, r=\frac{-1}{4}

\text{Second term=}ar=-2\times (\frac{-1}{4})=\frac{1}{2}

\text{Third term=}ar^2=-2\times (\frac{-1}{4})^2=\frac{-1}{8}

\text{Fourth term=}ar^3=-2\times (\frac{-1}{4})^3=\frac{1}{32}

\text{The next three terms of G.P is }\frac{1}{2}, \frac{-1}{8}, \frac{1}{32}

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3 years ago
A Martian couple has children until they have 2 males (sexes of children are independent). Compute the expected number of childr
Ne4ueva [31]

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a) 6

b) 4

c) 3

Step-by-step explanation:

Let p be the probability of having a female Martian, and of course, 1-p the probability of having a male Martian.

To compute the expected total number of trials before 2 males are born, imagine an experiment simulating the fact that 2 males are born is performed n times.

Let ak be the number of trials performed until 2 males are born in experiment k. That is,

a1= number of trials performed until 2 males are born in experiment 1

a2= number of trials performed until 2 males are born in experiment 2

and so on.

If a1 + a2 + … + an = N

we would expect Np females.  

Since the experiment was performed n times, there 2n males (recall that the experiment stops when 2 males are born).

So we would expect 2n = N(1-p), or

N/n = 2/(1-p)

But N/n is the average number of trials per experiment, that is, the expectation.

<em>We have then that the expected number of trials before 2 males are born is 2/(1-p) where p is the probability of having a female. </em>

a)

Here we have the probability of having a male is half as likely as females. So

1-p = p/2 hence p=2/3

The expected number of trials would be

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This means <em>the couple would have 6 children</em>: 4 females (the first 4 trials) and 2 males (the last 2 trials).

b)

Here the probability of having a female = probability of having a male = 1/2

The expected number of trials would be

2/(1/2) = 4

This means<em> the couple would have 4 children</em>: 2 females (the first 2 trials) and 2 males (the last 2 trials).

c)

Here, 1-p = 2p so p=1/3

The expected number of trials would be

2/(1-1/3) = 2/(2/3) = 6/2 =3

This means<em> the couple would have 3 children</em>: 1 female (the first trial) and 2 males (the last 2 trials).

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3 years ago
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