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Like how we have dealt with arithmetic progression, we will deal with geometric progression in this section.
Situation 1
In this situation, the midpoints of \(\triangle ABC\) is joined to form \(\triangle DEF\).
Similarly, the midpoint of \(\triangle CEF\) is joined to form \(\triangle MNO\).
The midpoints of \(\triangle CON\) is joined to form a different triangle and so on.
The area of \(\triangle ABC\), \(\triangle DEF\), \(\triangle MNO\) and so on is given by \(\triangle ABC\), \(\frac{1}{4}\triangle ABC\), \(\frac{1}{4} \times \frac{1}{4} \triangle ABC\) and so on.
That is, \(\triangle ABC\), \(\frac{1}{4} \triangle ABC\), \(\frac{1}{16} \triangle ABC\), and so on.
In other words, the areas of \(\triangle ABC\), \(\triangle DEF\), \(\triangle MNO\) and so on are \(\frac{1}{4}\) apart.
Therefore, the areas of the triangles formed are in a geometric progression with \(\frac{1}{4}\) as the common ratio.
Situation 2
A particular dog breed gives birth to exactly two puppies at a time.
The number of dogs in each level is indicated as \(1\), \(2\), \(4\),\(…\)
Therefore, the number of dogs is a geometric progression with \(2\) as the common ratio.
Geometric progression
A Geometric Progression is a sequence in which each term is obtained by multiplying a fixed non-zero number by the preceding term except the first term. The fixed number is called the common ratio. The common ratio is usually denoted by \(r\).
A geometric progression is given in the form of \(a\), \(ar\), \(ar^2\),\(...ar^{n-1}\).
Here, \(a\) is the first term, and \(r\) is the common ratio.
The first term \(a\), when multiplied by the \(r\) subsequently, forms \(a\), \(ar\), \(ar^2\),\(...ar^{n-1}\), which is the geometric progression.
let us look at the general formula to find the \(n^{th}\) term of a geometric progression.
We know that the general form of a geometric progression is \(a\), \(ar\), \(ar^2\),\(...ar^{n - 1}\) with the common ratio \(r\).
\(\text{Term } 1 = t_1 = a \times r^0 = a \times 1 = a\)
\(\text{Term } 2 = t_2 = a \times r^1 = ar\)
\(\text{Term } 3 = t_3 = a \times r^2 = ar^2\)
\(.\) \(.\)
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\(.\) \(.\)
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\(\text{Term } n = t_{n} = a \times r^{n - 1} = ar^{n - 1}\)
Explicit form: The general form of Geometric Progression is \( t_{n} = ar^{n - 1}\).
Recursive form: The general form of Geometric Progression is \( t_1=a,\quad t_n=rt_{n-1},\quad n\ge2 \)
Ratio between any two consecutive terms in a G.P.:
\(\frac{t_2}{t_1} = \frac{ar}{a} = r\)
\(\frac{t_3}{t_2} = \frac{ar^2}{ar} = r\)
\(\frac{t_n}{t_{n - 1}} = \frac{ar^n}{ar^{n - 1}} = \frac{ar^n}{a \times r^n \times r^{-1}} = \frac{1}{r{-1}} = r\)
Thus, the ratio between any two consecutive terms of a G.P. is \(r\).
Example:
1. Consider the sequence \(2,\ 6,\ 18,\ 54,\ 162,\ldots\). Determine whether it is a geometric progression. If so, find the common ratio and the \(n\text{th term.} \)
Given: \( t_1=2,\quad t_2=6,\quad t_3=18,\quad t_4=54 \)
To find: Find common ratio and \(n\text{th term}\)
\( \frac{t_2}{t_1}=\frac{6}{2}=3 \)
\( \frac{t_3}{t_2}=\frac{18}{6}=3 \)
\( \frac{t_4}{t_3}=\frac{54}{18}=3 \)
Therefore \(\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}=3 \)
Therefore the given sequence is a Geometric progression.
\( \frac{t_2}{t_1}=\frac{6}{2}=3 \)
\( \frac{t_3}{t_2}=\frac{18}{6}=3 \)
\( \frac{t_4}{t_3}=\frac{54}{18}=3 \)
Therefore \(\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}=3 \)
Therefore the given sequence is a Geometric progression.
To find the explicit formula:
\( a=2,\quad r=3 \)
\( t_n=ar^{n-1} \)
\( t_n=2(3)^{n-1} \)
\( a=2,\quad r=3 \)
\( t_n=ar^{n-1} \)
\( t_n=2(3)^{n-1} \)
To find the recursive formula:
\( t_1=2 \)
\( t_n=rt_{n-1},\quad n\ge2 \)
\( t_n=3t_{n-1},\quad n\ge2 \)
\( t_1=2 \)
\( t_n=rt_{n-1},\quad n\ge2 \)
\( t_n=3t_{n-1},\quad n\ge2 \)
2.Check whether \(486\) and \(500\) are terms of the G.P. \(2,\ 6,\ 18,\ 54,\ldots \)
Given : \( a=2,\quad r=3 \)
\( t_n=ar^{n-1} \)
\( t_n=ar^{n-1} \)
To check: \(486\) is a term of the sequence.
Explanation:
\( 486=2(3)^{n-1} \)
\( 3^{n-1}=243 \)
\( 3^{n-1}=3^5 \)
\( n-1=5 \)
\( n=6 \)
Therefore \(486=t_6 \)
\( 486\) is a term of the given \(G.P.\).
\( 486=2(3)^{n-1} \)
\( 3^{n-1}=243 \)
\( 3^{n-1}=3^5 \)
\( n-1=5 \)
\( n=6 \)
Therefore \(486=t_6 \)
\( 486\) is a term of the given \(G.P.\).
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