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ivann1987 [24]
3 years ago
5

A G.P is such that the third term is nine times the first term, while the second term is one twenty fourth of the fifth term fin

dits fourth term
Mathematics
1 answer:
denpristay [2]3 years ago
6 0

Answer:

Not a G.P

Step-by-step explanation:

Given - A G.P is such that the third term is nine times the first term, while the second term is one twenty fourth of the fifth term

To find - find its fourth term.

Proof -

Let first term of G.P = a

Second term of G.P = b

Third term of G.P = c

Fourth term of G.P = d

Fifth term of G.P = e

Now,

Given that the third term is nine times the first term

⇒c = 9a

And

Given that the second term is one twenty fourth of the fifth term

⇒b = \frac{1}{24} e

We know that in a g.P, there is common ratio in the terms

i.e \frac{b}{a} = \frac{c}{b} = \frac{d}{c} = \frac{e}{d} = r

and the nth term is represented as aₙ = arⁿ⁻¹

Now,

Second term would be b = ar

Third term would be c = ar²

fourth term would be d = ar³

Fifth term would be e = ar⁴

Now,

c = 9a

⇒ar² = 9a

⇒r² = 9

⇒r = 3 or -3

And

b = \frac{1}{24} e

⇒24b = e

⇒24(ar) = ar⁴

⇒24 = r³

As we have r = 3 or -3

S0 (3)³ = 27 ≠ 24

or (-3)³ = 27 ≠ 24

It is not possible.

So, it is not a G.P

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Answer:

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The p value for this case would be given by:

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Step-by-step explanation:

Information provided

\bar X=19.87 represent the sample mean

\sigma=0.4 represent the population deviation

n=25 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0.01 represent the significance level

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is at least 20 ounces, the system of hypothesis would be:  

Null hypothesis:\mu \geq 20  

Alternative hypothesis:\mu < 20  

The statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Replacing the info given we got:

z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625    

The p value for this case would be given by:

p_v =P(z  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

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