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Anna007 [38]
3 years ago
15

An infinite geometric series has 1 and 1/5 as its first two terms: 1, 1/5, 1/25, 1/125, . . . What is the sum, S, of the infinit

e series? A. 1 B. 5/4 C. 1/4 D. 1/25
Mathematics
1 answer:
dezoksy [38]3 years ago
7 0

Answer:  \dfrac{5}{4}

Step-by-step explanation:

Given: The first term of Geometric series : a=1

The seconds term of Geometric series : ar=\dfrac{1}{5}

The common ratio between the terms is given by :-

r=\dfrac{ar}{a}=\dfrac{\frac{1}{5}}{1}=\dfrac{1}{5}

We know that the sum of infinite geometric series is given by :-

S_{\infty}=\dfrac{a}{1-r}\\\\\Rightarrow\ S_{\infty}=\dfrac{1}{1-\frac{1}{5}}=\dfrac{1}{\frac{4}{5}}\\\\\Rightarrow\ S_{\infty}=\dfrac{5}{4}

Hence, the sum of the given infinite series = \dfrac{5}{4}

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\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
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\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
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\\ \quad \\
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\\\\
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\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{  D}}\\
\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\
\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind, let's see

\bf h(x)=-x+5\implies h(x)=\stackrel{A}{-1}(\stackrel{B}{1}x\stackrel{C}{+0})\stackrel{D}{+5}

a shift down by 3 units, means a vertical shift downwards, so D needs to drop by 3 units.

\bf h(x)=\stackrel{A}{-1}(\stackrel{B}{1}x\stackrel{C}{+0})\boxed{\stackrel{D}{+5-3}}\implies h(x)=-1(1x+0)+2
\\\\\\
h(x)=-x+2
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