Geometric Progression-Exercise 2

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– Posted on 24/03/2014Posted in: MATH

 

The sum of the numbers of an infinite decreasing geometric progression is 8.

Find y1 dhe q, considering that the sum of their cubes is \frac{512}{7}

Solution:

\frac{y_{1}}{1-q}=8

\frac{y_{1}^{3}}{1-q^{3}}=\frac{512}{7}

Make fraction in each side.

\frac{y_{1}^{3}}{1-q^{3}}\cdot \frac{1-q}{y_{1}}=\frac{512}{7}\cdot \frac{1}{8}=\frac{64}{7}

\frac{y_{1}^{2}}{1+q+q^{2}}=\frac{64}{7}

7y_{1}^{2}=64(1+q+q^{2})         y_{1}=8(1-q)

7\cdot 64(1-q)^{2}=64\cdot (1+q+q^{2})

7\cdot (1-2q+q^{2})= 1+q+q^{2}

7q^{2}-14q+7-1-q-q^{2}=0

6q^{2}-15q+6=0

2q^{2}-5q+2=0         D=25-16=9

_{2}q_{1}=\frac{5\pm 3}{4}                                                  q_{1}=2        q_{2}=\frac{1}{2}\rightarrowthe quotinent of decreasing geometriv progression(we don’t take q2, because then, it wont be an infinite decreasing geometric progression anymore.

q_{2}=\frac{1}{2}\Rightarrow y_{1}=8\cdot \frac{1}{2}=4

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