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Two distinct, real, infinite geometric series each have a sum of 1 and have the same second term. The third term of one of the series is 1/8. Find all possible values for the second term of both series.

 Sep 11, 2019

Best Answer 

 #1
avatar+23903 
+2

Two distinct, real, infinite geometric series each have a sum of 1 and have the same second term.

The third term of one of the series is 1/8. 

Find all possible values for the second term of both series.

 

1. geometric serie: \(s_1 = a+ar_1+ar_1^2+\ldots\)

2. geometric serie: \(s_2 = b+br_2+br_2^2+\ldots\)

 

The same second term: \( ar_1= br_2=x \)

 

Sum of 1:

\(\begin{array}{|rcll|} \hline s_1 = 1 &=& \dfrac{a}{1-r_1} \quad | \quad \Rightarrow \boxed{r_1=1-a} \\ s_2 = 1 &=& \dfrac{b}{1-r_2} \quad | \quad \Rightarrow \boxed{r_2=1-b\quad \text{or} \quad b=1-r_2 } \\ \hline \end{array}\)

 

The third term of one of the series is \(\dfrac{1}{8}\):

\(\begin{array}{|lrcll|} \hline & br_2^2 &=& \dfrac{1}{8} \\ & br_2r_2 &=& \dfrac{1}{8} \\ & r_2 &=& \dfrac{1}{8br_2} \quad | \quad br_2 = x \\ (1) & \mathbf{r_2} &=& \mathbf{\dfrac{1}{8x}} \quad | \quad r_2=1-b \\ & 1-b &=& \dfrac{1}{8x} \\ (2) &\mathbf{b} &=& \mathbf{1- \dfrac{1}{8x}} \\ \hline (2) \times(1) & \mathbf{br_2 } &=& \mathbf{\left(1- \dfrac{1}{8x} \right)\dfrac{1}{8x} } \quad | \quad br_2=x \\ & x &=& \left(1- \dfrac{1}{8x} \right)\dfrac{1}{8x} \\ & 8x^2 &=& 1- \dfrac{1}{8x} \quad | \quad \cdot 8x \\ & 64x^3 &=& 8x - 1 \\ &\mathbf{64x^3-8x+1} &=& \mathbf{0} \\\\ & \mathbf{x_1} &=& \mathbf{\dfrac{1}{4}} \\ & \mathbf{x_2} &=& \mathbf{-\dfrac{1}{8}\left( \sqrt{5} + 1 \right)} \\ & \mathbf{x_3} &=& \mathbf{\dfrac{1}{8}\left( \sqrt{5} - 1 \right)} \\ \hline \end{array}\)

 

Summary:

\(\begin{array}{|lrcll|} \hline & \mathbf{x_1} &=& \mathbf{\dfrac{1}{4}} \\ & \mathbf{x_2} &=& \mathbf{-\dfrac{1}{8}\left( \sqrt{5} + 1 \right)} \\ & \mathbf{x_3} &=& \mathbf{\dfrac{1}{8}\left( \sqrt{5} - 1 \right)} \\ \hline (1) & \mathbf{r_2} &=& \mathbf{\dfrac{1}{8x}} \\ (3) & \mathbf{b} &=& \mathbf{ 1-r_2 } \\ \hline & ar_1 &=& x \quad | \quad r_1 = 1-a \\ & a(1-a) &=& x \\ & a^2-a+x &=& 0 \\ (4) & \mathbf{a} &=& \mathbf{\dfrac{1\pm \sqrt{1-4x}}{2}} \\ (5) & \mathbf{r_1} &=& \mathbf{ 1-a } \\ \hline \end{array}\)

 

\(\begin{array}{|l|r|r|r|c|} \hline & r_2= & b= & a= & r_1= & \text{distinct} \\ \hline \mathbf{x=\dfrac{1}{4}} & \dfrac{1}{2} & \dfrac{1}{2} & \dfrac{1}{2} & \dfrac{1}{2} & \text{no} \\ \hline \mathbf{x=-\dfrac{1}{8}\left( \sqrt{5} + 1 \right)} & \dfrac{1-\sqrt{5}}{4} & \dfrac{3+\sqrt{5}}{4} & b & r_2 & \text{no} \\ & & & r_2 & b & \text{yes} \\ \hline \mathbf{x_3=\dfrac{1}{8}\left( \sqrt{5} - 1 \right)} & \dfrac{1+\sqrt{5}}{4} & \dfrac{3-\sqrt{5}}{4} & b & r_2 & \text{no} \\ & & & r_2 & b & \text{yes} \\ \hline \end{array} \)

 

Solution 1:

\(\begin{array}{|l|} \hline \mathbf{x=-\dfrac{1}{8}\left( \sqrt{5} + 1 \right)} \\ a=r_2 = \dfrac{1-\sqrt{5}}{4} \\ r_1 = b = \dfrac{3+\sqrt{5}}{4} \\ b = \dfrac{3+\sqrt{5}}{4} \\ r_2 = \dfrac{1-\sqrt{5}}{4} \\ \hline a+ar_1+ar_1^2+\ldots = \left( \dfrac{1-\sqrt{5}}{4} \right) +\underbrace{\left(\dfrac{1-\sqrt{5}}{4}\right)\left(\dfrac{3+\sqrt{5}}{4}\right)}_{=x} +\left(\dfrac{1-\sqrt{5}}{4}\right)\left(\dfrac{3+\sqrt{5}}{4}\right)^2 +\ldots \\\\ b+br_2+br_2^2+\ldots = \left( \dfrac{3+\sqrt{5}}{4} \right) +\overbrace{\left(\dfrac{3+\sqrt{5}}{4} \right)\left(\dfrac{1-\sqrt{5}}{4}\right)}^{=x} +\underbrace{\left(\dfrac{3+\sqrt{5}}{4} \right)\left(\dfrac{1-\sqrt{5}}{4}\right)^2}_{=\dfrac{1}{8}} +\ldots \\ \hline \end{array}\)

 

Solution 2:

\(\begin{array}{|l|} \hline x=\mathbf{\dfrac{1}{8}\left( \sqrt{5} - 1 \right)} \\ a=r_2 = \dfrac{1+\sqrt{5}}{4} \\ r_1 = b = \dfrac{3-\sqrt{5}}{4} \\ b = \dfrac{3-\sqrt{5}}{4} \\ r_2 = \dfrac{1+\sqrt{5}}{4} \\ \hline a+ar_1+ar_1^2+\ldots = \left( \dfrac{1+\sqrt{5}}{4} \right) +\underbrace{\left(\dfrac{1+\sqrt{5}}{4}\right)\left(\dfrac{3-\sqrt{5}}{4}\right)}_{=x} +\left(\dfrac{1+\sqrt{5}}{4}\right)\left(\dfrac{3-\sqrt{5}}{4}\right)^2 +\ldots \\\\ b+br_2+br_2^2+\ldots = \left( \dfrac{3-\sqrt{5}}{4} \right) +\overbrace{\left(\dfrac{3-\sqrt{5}}{4} \right)\left(\dfrac{1+\sqrt{5}}{4}\right)}^{=x} +\underbrace{\left(\dfrac{3-\sqrt{5}}{4} \right)\left(\dfrac{1+\sqrt{5}}{4}\right)^2}_{=\dfrac{1}{8}} +\ldots \\ \hline \end{array}\)

 

Find all possible values for the second term of both series:

\(\begin{array}{|rcll|} \hline \text{The second term: } & \left(\dfrac{1-\sqrt{5}}{4}\right)\left(\dfrac{3+\sqrt{5}}{4}\right)&=&-\dfrac{1}{8}\left( \sqrt{5} + 1 \right) \\ \text{The second term: } & \left(\dfrac{1+\sqrt{5}}{4}\right)\left(\dfrac{3-\sqrt{5}}{4}\right)&=&\dfrac{1}{8}\left( \sqrt{5} - 1 \right) \\ \hline \end{array} \)

 

laugh

 Sep 12, 2019
 #1
avatar+23903 
+2
Best Answer

Two distinct, real, infinite geometric series each have a sum of 1 and have the same second term.

The third term of one of the series is 1/8. 

Find all possible values for the second term of both series.

 

1. geometric serie: \(s_1 = a+ar_1+ar_1^2+\ldots\)

2. geometric serie: \(s_2 = b+br_2+br_2^2+\ldots\)

 

The same second term: \( ar_1= br_2=x \)

 

Sum of 1:

\(\begin{array}{|rcll|} \hline s_1 = 1 &=& \dfrac{a}{1-r_1} \quad | \quad \Rightarrow \boxed{r_1=1-a} \\ s_2 = 1 &=& \dfrac{b}{1-r_2} \quad | \quad \Rightarrow \boxed{r_2=1-b\quad \text{or} \quad b=1-r_2 } \\ \hline \end{array}\)

 

The third term of one of the series is \(\dfrac{1}{8}\):

\(\begin{array}{|lrcll|} \hline & br_2^2 &=& \dfrac{1}{8} \\ & br_2r_2 &=& \dfrac{1}{8} \\ & r_2 &=& \dfrac{1}{8br_2} \quad | \quad br_2 = x \\ (1) & \mathbf{r_2} &=& \mathbf{\dfrac{1}{8x}} \quad | \quad r_2=1-b \\ & 1-b &=& \dfrac{1}{8x} \\ (2) &\mathbf{b} &=& \mathbf{1- \dfrac{1}{8x}} \\ \hline (2) \times(1) & \mathbf{br_2 } &=& \mathbf{\left(1- \dfrac{1}{8x} \right)\dfrac{1}{8x} } \quad | \quad br_2=x \\ & x &=& \left(1- \dfrac{1}{8x} \right)\dfrac{1}{8x} \\ & 8x^2 &=& 1- \dfrac{1}{8x} \quad | \quad \cdot 8x \\ & 64x^3 &=& 8x - 1 \\ &\mathbf{64x^3-8x+1} &=& \mathbf{0} \\\\ & \mathbf{x_1} &=& \mathbf{\dfrac{1}{4}} \\ & \mathbf{x_2} &=& \mathbf{-\dfrac{1}{8}\left( \sqrt{5} + 1 \right)} \\ & \mathbf{x_3} &=& \mathbf{\dfrac{1}{8}\left( \sqrt{5} - 1 \right)} \\ \hline \end{array}\)

 

Summary:

\(\begin{array}{|lrcll|} \hline & \mathbf{x_1} &=& \mathbf{\dfrac{1}{4}} \\ & \mathbf{x_2} &=& \mathbf{-\dfrac{1}{8}\left( \sqrt{5} + 1 \right)} \\ & \mathbf{x_3} &=& \mathbf{\dfrac{1}{8}\left( \sqrt{5} - 1 \right)} \\ \hline (1) & \mathbf{r_2} &=& \mathbf{\dfrac{1}{8x}} \\ (3) & \mathbf{b} &=& \mathbf{ 1-r_2 } \\ \hline & ar_1 &=& x \quad | \quad r_1 = 1-a \\ & a(1-a) &=& x \\ & a^2-a+x &=& 0 \\ (4) & \mathbf{a} &=& \mathbf{\dfrac{1\pm \sqrt{1-4x}}{2}} \\ (5) & \mathbf{r_1} &=& \mathbf{ 1-a } \\ \hline \end{array}\)

 

\(\begin{array}{|l|r|r|r|c|} \hline & r_2= & b= & a= & r_1= & \text{distinct} \\ \hline \mathbf{x=\dfrac{1}{4}} & \dfrac{1}{2} & \dfrac{1}{2} & \dfrac{1}{2} & \dfrac{1}{2} & \text{no} \\ \hline \mathbf{x=-\dfrac{1}{8}\left( \sqrt{5} + 1 \right)} & \dfrac{1-\sqrt{5}}{4} & \dfrac{3+\sqrt{5}}{4} & b & r_2 & \text{no} \\ & & & r_2 & b & \text{yes} \\ \hline \mathbf{x_3=\dfrac{1}{8}\left( \sqrt{5} - 1 \right)} & \dfrac{1+\sqrt{5}}{4} & \dfrac{3-\sqrt{5}}{4} & b & r_2 & \text{no} \\ & & & r_2 & b & \text{yes} \\ \hline \end{array} \)

 

Solution 1:

\(\begin{array}{|l|} \hline \mathbf{x=-\dfrac{1}{8}\left( \sqrt{5} + 1 \right)} \\ a=r_2 = \dfrac{1-\sqrt{5}}{4} \\ r_1 = b = \dfrac{3+\sqrt{5}}{4} \\ b = \dfrac{3+\sqrt{5}}{4} \\ r_2 = \dfrac{1-\sqrt{5}}{4} \\ \hline a+ar_1+ar_1^2+\ldots = \left( \dfrac{1-\sqrt{5}}{4} \right) +\underbrace{\left(\dfrac{1-\sqrt{5}}{4}\right)\left(\dfrac{3+\sqrt{5}}{4}\right)}_{=x} +\left(\dfrac{1-\sqrt{5}}{4}\right)\left(\dfrac{3+\sqrt{5}}{4}\right)^2 +\ldots \\\\ b+br_2+br_2^2+\ldots = \left( \dfrac{3+\sqrt{5}}{4} \right) +\overbrace{\left(\dfrac{3+\sqrt{5}}{4} \right)\left(\dfrac{1-\sqrt{5}}{4}\right)}^{=x} +\underbrace{\left(\dfrac{3+\sqrt{5}}{4} \right)\left(\dfrac{1-\sqrt{5}}{4}\right)^2}_{=\dfrac{1}{8}} +\ldots \\ \hline \end{array}\)

 

Solution 2:

\(\begin{array}{|l|} \hline x=\mathbf{\dfrac{1}{8}\left( \sqrt{5} - 1 \right)} \\ a=r_2 = \dfrac{1+\sqrt{5}}{4} \\ r_1 = b = \dfrac{3-\sqrt{5}}{4} \\ b = \dfrac{3-\sqrt{5}}{4} \\ r_2 = \dfrac{1+\sqrt{5}}{4} \\ \hline a+ar_1+ar_1^2+\ldots = \left( \dfrac{1+\sqrt{5}}{4} \right) +\underbrace{\left(\dfrac{1+\sqrt{5}}{4}\right)\left(\dfrac{3-\sqrt{5}}{4}\right)}_{=x} +\left(\dfrac{1+\sqrt{5}}{4}\right)\left(\dfrac{3-\sqrt{5}}{4}\right)^2 +\ldots \\\\ b+br_2+br_2^2+\ldots = \left( \dfrac{3-\sqrt{5}}{4} \right) +\overbrace{\left(\dfrac{3-\sqrt{5}}{4} \right)\left(\dfrac{1+\sqrt{5}}{4}\right)}^{=x} +\underbrace{\left(\dfrac{3-\sqrt{5}}{4} \right)\left(\dfrac{1+\sqrt{5}}{4}\right)^2}_{=\dfrac{1}{8}} +\ldots \\ \hline \end{array}\)

 

Find all possible values for the second term of both series:

\(\begin{array}{|rcll|} \hline \text{The second term: } & \left(\dfrac{1-\sqrt{5}}{4}\right)\left(\dfrac{3+\sqrt{5}}{4}\right)&=&-\dfrac{1}{8}\left( \sqrt{5} + 1 \right) \\ \text{The second term: } & \left(\dfrac{1+\sqrt{5}}{4}\right)\left(\dfrac{3-\sqrt{5}}{4}\right)&=&\dfrac{1}{8}\left( \sqrt{5} - 1 \right) \\ \hline \end{array} \)

 

laugh

heureka Sep 12, 2019

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