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Yuri [45]
4 years ago
9

Find the sum of the first five terms using the geometric series formula for the sequence left curly bracket 1 half comma space m

inus 1 comma 2 comma negative 4 comma... right curly bracket
Mathematics
1 answer:
MatroZZZ [7]4 years ago
3 0

Answer:

\bold{\dfrac{11}{2 }}

Step-by-step explanation:

Given the geometric series:

\{\dfrac{1}2, -1, 2, -4,  ..... \}

To find:

Sum of series upto 5 terms using the geometric series formula = ?

Solution:

Formula for sum of a n terms of a geometric series is given as:

S_n=\dfrac{a(1-r^n)}{1-r} \ \{r

a is the first term of the geometric series

r is the common ratio between each term (2nd term divided by 1st term or 3rd term divided by 2nd term ..... ).

Here:

a = \dfrac{1}{2}

r = \dfrac{-1}{\dfrac{1}{2}} = -2

n=5

So, applying the formula for given values:

S_5=\dfrac{\dfrac{1}2(1-(-2)^5)}{1-(-2)} \\\Rightarrow S_5=\dfrac{1-(-32)}{2 \times 3} \\\Rightarrow S_5=\dfrac{1+32}{6} \\\Rightarrow S_5=\dfrac{33}{6} \\\Rightarrow \bold{S_5=\dfrac{11}{2}}

So, the answer is

\bold{\dfrac{11}{2 }}

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t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28    

The degrees of freedom are given by

df=n-1=17-1=16  

The p value would be given by this probability:

p_v =P(t_{(16)}  

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Step-by-step explanation:

Information given

\bar X=8.19 cm^3 represent the sample mean

s=0.8 cm^3 represent the sample deviation

n=17 sample size  

\mu_o =9.02 represent the value to verify

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic

p_v represent the p value

System of hypothesis to check

We want to check if the true mean is less than the normal value of 9.02 cm^3, the system of hypothesis would be:  

Null hypothesis:\mu \geq 9.02  

Alternative hypothesis:\mu < 9.02  

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28    

The degrees of freedom are given by

df=n-1=17-1=16  

The p value would be given by this probability:

p_v =P(t_{(16)}  

Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3

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