# 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.

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

$P_{1}=4a$

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

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

We verify that this string is geometric progression.

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

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

$q=\frac{\sqrt{2}}{2}<&space;1\Rightarrow$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&space;\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.

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

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

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

We find the quotient

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

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

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

Then we find the sum

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