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jenyasd209 [6]
3 years ago
10

In an arithmetic​ sequence, the nth term an is given by the formula an=a1+(n−1)d​, where a1 is the first term and d is the commo

n difference.​ Similarly, in a geometric​ sequence, the nth term is given by 1an=a1•rn−1​, where r is the common ratio. Use these formulas to determine the indicated term in the given sequence.
The 10th term of 40,10, 5/2, 5/8, ....
Mathematics
1 answer:
Dmitry_Shevchenko [17]3 years ago
3 0

Answer:

a_{10} = \frac{10}{65536}

Step-by-step explanation:

The first step to solving this problem is verifying if this sequence is an arithmetic sequence or a geometric sequence.

This sequence is arithmetic if:

a_{3} - a_{2} = a_{2} - a_{1}

We have that:

a_{3} = 40, a_{2} = 10, a_{3} = \frac{5}{2}

a_{3} - a_{2} = a_{2} - a_{1}

\frac{5}{2} - 10 = 10 - 40

\frac{-15}{2} \neq -30

This is not an arithmetic sequence.

This sequence is geometric if:

\frac{a_{3}}{a_{2}} = \frac{a_{2}}{a_{1}}

\frac{\frac{5}[2}}{10} = \frac{10}{40}

\frac{5}{20} = \frac{1}{4}

\frac{1}{4} = \frac{1}{4}

This is a geometric sequence, in which:

The first term is 40, so a_{1} = 40

The common ratio is \frac{1}{4}, so r = \frac{1}{4}.

We have that:

a_{n} = a_{1}*r^{n-1}

The 10th term is a_{10}. So:

a_{10} = a_{1}*r^{9}

a_{10} = 40*(\frac{1}{4})^{9}

a_{10} = \frac{40}{262144}

Simplifying by 4, we have:

a_{10} = \frac{10}{65536}

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17x-3y=4 2x-4y=1 the solution to the system of equations is
Burka [1]

ANSWER

x=\frac{13}{62}

and


y= \frac{-9}{62}e have



EXPLANATION




17x-3y=4---(1)


2x-4y=1



Let us make y the subject and call it equation (2)


2x=1+4y


x=\frac{1}{2}+2y--(2)



We put equation (2) in to equation (1)




17(\frac{1}{2}+2y)-3y=4


\frac{17}{2}+34y-3y=4



34y-3y=4- \frac{17}{2}


Simplify to get,



31y= \frac{8-17}{2}



31y= \frac{-9}{2}


Divide both sides by 31,



y= \frac{-9}{2} \div 31



y= \frac{-9}{2} \times \frac{1}{31}




y= \frac{-9}{62}


We put this value in to equation (2) to get,



x=\frac{1}{2}+2\times -\frac{9}{62}




x=\frac{1}{2} -\frac{18}{62}


We collect LCM to obtain,



x=\frac{31-18}{62}



x=\frac{13}{62}








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