The geometric progression is type of progression in which the quotient of every member and the previous member is constant i.e. \(q = \frac{a_n}{a_{n-1}}, n>1.\) The first member of the geometric progression is labeled as \(a_1\), the second as \(a_2 = a_1\cdot q\), the third as \(a_3 = a_2\cdot q\) or \(a_3 = a_1 \cdot q^2\), generally the nth member of geometric progression can be determined from formula:
$$ a_n = a_{n-1} \cdot q \quad \mathrm{or} \quad a_n = a_1 \cdot q^{n-1}.$$
The sum of first \(n\) members can be calculated using formula:
\begin{eqnarray}
S_n &=& a_1\frac{q^{n}-1}{q-1}, \quad q\neq 1.
\end{eqnarray}
For every three consecutive members of geometric progression the following formula is valid.
\begin{eqnarray}
\frac{a_n}{a_{n-1}} &=& \frac{a_{n+1}}{a_n} \Rightarrow a_1 = \sqrt{a_{n-1}\cdot a_{n+1}}
\end{eqnarray}
The geometric average of \(n\) numbers can be calculated as:
$$ G = \sqrt[n]{a_1 \cdot a_2 \cdots a_n }.$$
The inserted (interpolated) geometric sequence that consists of \(r\) members between numbers \(a\) and \(b\) means to determine \(r\) numbers that together with \(a\) and \(b\) form geometric sequence. The number \(a\) is the first member while \(b\) is the last member of the sequence. Quotient \(q\) between sequence members can be determine from
\begin{eqnarray}
a_1 &=& a, \\
a_n &=& b,\\
n &=& r +2, \\
a_n &=& a_1\cdot q^{n-1} \Rightarrow b = aq^{r+1} \\
q &=& \sqrt[r+1]{\frac{b}{a}}
\end{eqnarray}
- Example 1 Determine the sum of first five members of geometric progression 8,4,2,...
SolutionThe first member of the sequence is 8 i.e. \(a_1 = 8\) while the second member of the sequence is 4 i.e. \(a_2 = 4\). The quotient can be calculated as $$ q = \frac{a_2}{a_1} = \frac{4}{8} = \frac{1}{2}. $$ The sum of first 5 members of geometric progression is \begin{eqnarray} S_5 &=& a_1\frac{q^{5}-1}{q-1}\\ S_5 &=& 8\frac{\left(\frac{1}{2}\right)^5 - 1}{\left(\frac{1}{2}\right) -1} \\ S_5 &=& 8\frac{\frac{1-32}{32}}{\frac{1-2}{2}}\\ S_5 &=& 8\frac{\frac{31}{32}}{\frac{1}{2}} = 8\frac{31}{16} = \frac{31}{2}\\ \end{eqnarray} - Example 2 Determine the geometric progression in which the following equations are valid \(a_4 - a_2 = 120\) and \(a_1\cdot a_3 = 64.\)
Solution The first four members of geometric progression are \(a_1, a_2 = a_1\cdot q\), \(a_3 = a_1\cdot q^2,\) and \(a_4=a_1\cdot q^3.\) Inserting them into initial equations we can find the members of geometric sequence. \begin{eqnarray} a_4 - a_2 &=& 120 \Rightarrow a_1q^3 - a_1q = 120 \Rightarrow a_1q(q^2-1)=120 \\ a_1\cdot a_1\cdot q^2 &=& 64 \Rightarrow a_1^2 q^2 = 64/\sqrt \Rightarrow a_1q = \pm 8️ \end{eqnarray} By dividing those two equations we obtain: \begin{eqnarray} \frac{a_1q(q^2-1)}{a_1q} &=& \frac{120}{\pm 8}\\ q^2 -1 &=& \pm 15\\ q^2 &=& \pm 15 + 1 \\ q_1^2 &=& 16 \Rightarrow q_1 = 4 \\ q_2^2 &=& -14 \Rightarrow \mathrm{No}\quad\mathrm{solution}\\ a_1q &=& \pm 8 \Rightarrow 4a_1 = \pm 8/:4 \Rightarrow a_1 =\pm 2 \end{eqnarray} The two solutions i.e. geometric progressions can be written as: I sequence: \(a_1 = 2, q = 4 \) II sequence: \(a_1 = -2, q= -4\) - Example 3 The sum of first three members of growing geometric sequence is 63, and the product is 1728. Determine that sequence.
Solution \begin{eqnarray} a_1 + a_2 + a_3 &=& 63\\ a_1a_2a_3 &=& 1728 \end{eqnarray} The first, second and third member of the sequence can be written as: \begin{eqnarray} a_1 &=& a_1\\ a_2 &=& a_1 \cdot q \\ a_3 &=& a_1 \cdot q^2 \end{eqnarray} Substituting these expressions into the equations the following is obtained: \begin{eqnarray} a_1 + a_1q + a_1q^2 &=& 63\\ a_1^3q^3 &=& 1728/\sqrt[3]{}\\ a_1q = 12 \Rightarrow a_1 = \frac{12}{q}\\ \frac{12}{q} + 12 + 12q &=& 63/\cdot q\\ 12 + 12q + 12q^2 &=& 63q \\ 12q^2 - 51q + 12 &=& 0\\ \end{eqnarray} \begin{eqnarray} q_{1,2} &=& \frac{51 \pm \sqrt{2601- 576}}{24}\\ q_{1,2} &=& \frac{51 \pm 45}{24}\\ q_1 &=& \frac{51+45}{24} = 4 \\ q_2 &=& \frac{1}{4}\\ q_1 &=& 4 \Rightarrow a_1 = \frac{12}{q_1} = \frac{12}{4} = 3 \\ a_2 &=& 3\cdot 4 = 12, a_3 = 3 \cdot 16 = 48 \\ q_2 &=& \frac{1}{4}\Rightarrow a_1 = \frac{12}{q_2} = \frac{12}{\frac{1}{4}} = 48\\ a_2 &=& 48 \cdot \frac{1}{4} = 12, a_3 = 48 \cdot \frac{1}{16} = 3 \end{eqnarray} - Example 4 The first member of geometric sequence is \(a_1=-\frac{1}{2}\), and the sum of first three members is \(S_3 = -\frac{3}{2}\). Calculate the fifth member.
Solution \begin{eqnarray} a_1 &=& -\frac{1}{2}\\ S_3 = a_1\frac{q^3-1}{q-1}\\ -\frac{1}{2}(q^2 - q + 1) &=& -\frac{3}{2}/\cdot(-2)\\ q^2 - q - 2 &=& 0\\ q_{1,2} &=& \frac{1\pm \sqrt{1+8}}{2} = \frac{1 \pm 3}{2}\\ q_1 &=& \frac{4}{2} = 2 \\ q_2 &=& \frac{-2}{2} = -1\\ \end{eqnarray} The \(q_1 = 2\) is the quotient of geometric progression. The value of fifth member is: \begin{eqnarray} q_1 &=& 2 \Rightarrow a_5 = a_1\cdot q^4\\ a_5 &=& -\frac{1}{2}16 = -8 \end{eqnarray} - Example 5 The number of bacteria is doubles every two hours. At the beginning there was 1000 bacteria calculate the number of bacteria after 96 hours.
Solution At the beginning we have \(a_1 = 1000\) bacteria and after two hours we have \(a_2 = 1000\cdot 2 = 2000\) bacteria. After four hours we have \(a_3 = 2000 \cdot 2 = 4000\) bacteria. Based on the obtained results the general formula can be written as \(a_n = a_1 \cdot 2^{n-1}\). The equation is mathematical description of geometric progression. The next step is to determine the \(n\) after 96 hours. \(n = \frac{96}{2} + 1 = 49\). The number of bacteria after 96 hours can be calculated as: \begin{eqnarray} a_{49} &=& a_1 \cdot 2^{49-1} = 1000 \cdot 2^{48}\\ a_{49} &=& 1000\cdot 1024 \cdot 1024 \cdot 1024 \cdot 1024 \cdot 512\\ a_{49} &=& 281474976710656000 \end{eqnarray} - Example 6 The first member of geometric progression is 9 while third member is 121. Calculate the second member of the progression.
Solution The first step is to determine the quotient \(q\), and after the quotient is determine we can calculate the second member of geometric progression. \begin{eqnarray} a_3 &=& a_1\cdot q^2\\ 121 &=& 9\cdot q^2/:9 \\ q^2 &=& \frac{121}{9}/\sqrt{} \Rightarrow q = \pm\frac{11}{3} a_2 &=& a_1 \cdot q \\ a_2 &=& 9\cdot \pm\frac{11}{3} \\ a_2 &=& \pm33 \end{eqnarray} - Example 7 The geometric progression is 24, 6, 1.5,... Calculate the quotient.
Solution \begin{eqnarray} a_2 &=& a_1\cdot q\\ 6 &=& 24q/:24\\ q &=& \frac{1}{4} \end{eqnarray} To test if the quotient value is correct let's calculate the a_3. \begin{eqnarray} a_3 &=& a_1 \cdot q^2 \\ a_3 &=& 24 \cdot \left(\frac{1}{4}\right)^2\\ a_3 &=& 24 \cdot \frac{1}{16}\\ a_3 &=& \frac{3}{2} \end{eqnarray} - Example 8 Calculate the next three terms of the following geometric progression 2,40, 800,...
Solution \begin{eqnarray} a_2 &=& a_1 \cdot q\\ 40 &=& 2q /:2 \\ q &=& 20\\ a_4 &=& a_3\cdot q \\ a_4 &=& 800 \cdot 20 \\ a_4 &=& 16000\\ a_5 &=& a_4 \cdot q \\ a_5 &=& 16000 \cdot 20 \Rightarrow a_5 = 320000\\ a_6 &=& a_5 \cdot q \\ a_6 &=& 320000 \cdot 20 \Rightarrow a_6 = 6400000 \end{eqnarray} - Example 9 Calculate the next three members of the following geometric progression 75,15,3,...
Solution \begin{eqnarray} a_2 &=& a_1q \\ 15 &=& 75q/:75\\ q &=& \frac{1}{5}\\ a_4 &=& a_3q \\ a_4 &=& 3\cdot \frac{1}{5}\Rightarrow a_4 = 0.6\\ a_5 &=& 0.6 \cdot \frac{1}{5}\Rightarrow a_5 = 0.12\\ a_6 &=& 0.12 \cdot \frac{1}{5} \Rightarrow a_6 = 0.024 \end{eqnarray} - Example 10 Calculate the twentieth member of geometric sequence 2, 10, 50,...
Solution \begin{eqnarray} a_2 &=& a_1q \\ 10 &=& 2q /:2 \\ q &=& \frac{10}{2} \Rightarrow q = 5\\ a_{20} &=& a_1 q^{19} \\ a_{20} &=& 2 \cdot 5^{19} \\ a_{20} &=& 2 \cdot 3125 \cdot 3125 \cdot 3125 \cdot 625\\ a_{20} &=& 38146972656250 \end{eqnarray}
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