Geometric progression-Exercise

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– Posted on 24/03/2014Posted in: MATH

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\Rightarrowthe 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}

 

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