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# HELP!

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A geometric series $$b_1+b_2+b_3+\cdots+b_{10}$$ has a sum of 180. Assuming that the common ratio of that series is 7/4, find the sum of the series $$b_2+b_4+b_6+b_8+b_{10}.$$

Jun 17, 2019
edited by Guest  Jun 17, 2019

#1
+25568
+5

A geometric series $$b_1+b_2+b_3+\cdots+b_{10}$$ has a sum of 180.
Assuming that the common ratio of that series is $$\dfrac{7}{4}$$,
find the sum of the series $$b_2+b_4+b_6+b_8+b_{10}$$.

My attempt:

$$\text{Let b_2+b_4+b_6+b_8+b_{10} = x }$$

$$\begin{array}{|lrcll|} \hline & b_1+b_2+b_3+b_4+b_5+b_6+b_7+b_8+b_9+b_{10} &=& 180 \\ & (b_1+ b_3+ b_5+ b_7+ b_9)+(b_2+b_4+b_6+b_8+b_{10}) &=& 180 \\ & (b_1+ b_3+ b_5+ b_7+ b_9)+x &=& 180 \\ (1) & \mathbf{ b_1+ b_3+ b_5+ b_7+ b_9} &=& \mathbf{180 -x} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \text{common ratio }=\dfrac{7}{4} = \dfrac{b_2}{b_1}= \dfrac{b_6}{b_5}= \dfrac{b_8}{b_7}=\dfrac{b_{10}}{b_9}=\tan(\varphi) \quad | \quad =\text{ slope of the red line} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline &\text{slope of the red line} = \text{common ratio} = \dfrac{7}{4} &=& \dfrac{b_2+b_4+b_6+b_8+b_{10}}{b_1+ b_3+ b_5+ b_7+ b_9} \\ & \dfrac{7}{4} &=& \dfrac{x}{b_1+ b_3+ b_5+ b_7+ b_9} \\ (2)& \mathbf{b_1+ b_3+ b_5+ b_7+ b_9} &=& \mathbf{\dfrac{4}{7}x} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (1) & \mathbf{ b_1+ b_3+ b_5+ b_7+ b_9} &=& \mathbf{180 -x} \\ (2)& \mathbf{b_1+ b_3+ b_5+ b_7+ b_9} &=& \mathbf{\dfrac{4}{7}x} \\ \hline & 180 -x &=& \dfrac{4}{7}x \quad | \quad \cdot 7 \\ & 7\cdot 180 -7x &=& 4x \\ & 11x &=& 7\cdot 180 \\\\ & x &=& \dfrac{7\cdot 180}{11} \\ \\ & \mathbf{x} &=& \mathbf{ \dfrac{1260}{11} } \\ \hline \end{array}$$

The sum of the series $$b_2+b_4+b_6+b_8+b_{10} = \mathbf{ \dfrac{1260}{11} }$$.

Jun 18, 2019
edited by heureka  Jun 18, 2019
#4
+111464
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That's a very interesting approach, heureka  !!!!

I like it  !!!!

CPhill  Jun 18, 2019
#6
+25568
+2

Thank you, CPhill !

heureka  Jun 20, 2019
#2
+31085
+4

Different approach, same result:

.

Jun 18, 2019
#3
0

solve 180 = F×(1 - 1.75^10)/(1 - 1.75) for F
First term=0.503......
∑[0.503 * 1.75^(2n - 1), n , 1, 5] =1260 / 11

Jun 18, 2019
#5
+140
+5

Yet another different way of looking at it.

$$\displaystyle b_{1}+b_{2}+\dots+b_{10}=\frac{b_{1}[(7/4)^{10}-1]}{(7/4)-1}=180,\dots \dots(1)\\ \text{ so }\\b_{1}[(7/4)^{10}-1]=180.\frac{3}{4}=135.$$

Now consider the "same" GP but with a common ratio of -7/4.

$$\displaystyle b_{1}-b_{2}+b_{3}- \dots -b_{10}=\frac{b_{1}[(-7/4)^{10}-1]}{(-7/4)-1}=\frac{135}{(-11/4)}=-\frac{540}{11}\dots \dots(2)$$

Subtract (2) from (1),

$$\displaystyle 2(b_{2}+b_{4}+\dots +b_{10})=180+\frac{540}{11}=\frac{2520}{11},\\ \text{ so }\\ \displaystyle b_{2}+b_{4}+\dots +b_{10}=\frac{1260}{11}.$$

Jun 19, 2019
#7
+111464
+1

Nice....Tiggsy  and Alan   !!!!

CPhill  Jun 20, 2019