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Eduardwww [97]
1 year ago
8

THE SUM 10 INFINITY OF A GEOMETRIC SERIES IS 100. FIND THE TERM IF THE COMMON FIRST RATIO IS - 1/2​

Mathematics
1 answer:
zhuklara [117]1 year ago
6 0

Answer:

a = 150

Step-by-step explanation:

<u>Sum to infinity of a geometric series</u>

S_{\infty}=\dfrac{a}{1-r}\:\textsf{ for }|r| < 1

where:

  • a is the first term
  • r is the common ratio

Given:

  • S_{\infty}=100
  • r=-\dfrac{1}{2}

Substitute the given values into the formula and solve for a:

\begin{aligned}S_{\infty} & =\dfrac{a}{1-r}\\\\\implies 100 & =\dfrac{a}{1-\left(-\frac{1}{2}\right)}\\\\ 100 & =\dfrac{a}{1+\frac{1}{2}}\\\\100 & = \dfrac{a}{\frac{3}{2}}\\\\a & = \dfrac{3}{2} \cdot 100\\\\a & = 150\end{aligned}

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Explanation:

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Given the original function:  

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</span>→  <span>Write the original function as:  " y = </span>(5/9) (x − 32) " ; 

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    x = (5/9) (y − 32) ; 

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→ That is, solve this equation for "y" ; with "c" as an "isolated variable" on the
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Note the "distributive property" of multiplication:
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a(b + c) = ab + ac ;  <u><em>AND</em></u>:

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As such:
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" (\frac{5}{9}) * (y − 32) " ; 

=  [ (\frac{5}{9}) * y ]   −  [ (\frac{5}{9}) * (32) ] ; 


=  [ (\frac{5}{9}) y ]  − [ (\frac{5}{9}) * (\frac{32}{1})" ;

=  [ (\frac{5}{9}) y ]  − [ (\frac{(5*32)}{(9*1)} ] ; 

=  [ (\frac{5}{9}) y ]  −  [ (\frac{(160)}{(9)} ] ; 

= [ (\frac{5y}{9}) ]  −  [ (\frac{(160)}{(9)} ] ; 

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