Geometric progression-Exercise

Exercise 1

In a square that has side a, by bringing together the middle of the sides,is formed a new square.

In the new square ,in the same way,is formed another new square,and by the same wasy another one… infinity

a) Show that the string of perimetres of the squares is infinite decreasing geometric progression.

$P_{1} ; P_{2} ; P_{3}\rightarrow$ geometric progression

$P_{1}=4a$

$P_{2}=4\cdot A_{1}B_{1}=4\cdot \frac{a\sqrt{2}}{2}=2a\sqrt{2}$

$P_{3}=4\cdot A_{2}B_{2}=4\cdot \frac{a}{2}=2a$

We verify that this string is geometric progression.

$\frac{P_{2}}{P_{1}}=\frac{2a\sqrt{2}}{4a}=\frac{\sqrt{2}}{2}< 1$

$\frac{P_{3}}{P_{2}}=\frac{2a}{2a\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

$q=\frac{\sqrt{2}}{2}< 1\Rightarrow$ the string of perfimetres of the squares is infinite decreasing geometric progression.

a) Find the sum of the perimetres of the squares

$S=\frac{y_{1}}{1-q}=\frac{4a}{1-\frac{\sqrt{2}}{2}}=4a\cdot \frac{2}{2-\sqrt{2}}=\frac{8a}{2-\sqrt{2}}=\frac{8a(2+\sqrt{2})}{4-2}$ $=4a(2+\sqrt{2})$

Find the sum of the areas of the squares.

$\bullet a_{1}=a\Rightarrow S_{1}=a^{2}$

$\bullet a_{2}=\frac{a\sqrt{2}}{2}\Rightarrow S_{2}=\frac{2a^{2}}{4}=\frac{a^{2}}{2}$

$\bullet a_{3}=\frac{a}{2}\Rightarrow S_{3}=\frac{a^{2}}{4}$

We find the quotient

$\frac{S_{2}}{S_{1}}=\frac{\frac{a^{2}}{2}}{a^{2}}=\frac{a^{2}}{2}\cdot \frac{1}{a^{2}}=\frac{1}{2}$

$\frac{S_{3}}{S_{2}}=\frac{\frac{a^{2}}{4}}{\frac{a^{2}}{2}}=\frac{a^{2}}{4}\cdot \frac{2}{a^{2}}=\frac{1}{2}$

$q=\frac{1}{2}$

Then we find the sum

$S=\frac{y_{1}}{1-q}=\frac{a^{2}}{1-\frac{1}{2}}=2a^{2}$

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.

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

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

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

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

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

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

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

$2q^{2}-5q+2=0$ $D=25-16=9$

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

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

MATH

Geometric progression-Exercise

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