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A Geometric Progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio.
If the first term of a GP is \( a \) and the common ratio is \( r \), then the sequence is:
\( a, \quad ar, \quad ar^2, \quad ar^3, \quad \ldots \)
The nth term of a GP, denoted by \( T_n \), is given by the formula:
\( T_n = a r^{n-1} \)
Note: The common ratio \( r \) can be positive, negative, or fractional, but \( r \neq 0 \).
The sum of the first \( n \) terms of a GP is denoted by \( S_n \). The formula depends on the value of the common ratio \( r \):
\( S_n = a + ar + ar^2 + \ldots + ar^{n-1} = a \frac{1 - r^n}{1 - r} \)
\( S_n = n \times a \)
When the number of terms tends to infinity, the sum of the GP converges only if the common ratio satisfies:
\( |r| < 1 \)
In this case, the sum to infinity \( S_\infty \) is:
\( S_\infty = \frac{a}{1 - r} \)
If \( |r| \geq 1 \), the sum to infinity does not exist (it diverges).
Below is a simple diagram illustrating the first five terms of a GP with \( a = 2 \) and \( r = 3 \):
Term index (n): 1 2 3 4 5Term value: 2 6 18 54 162
Notice how each term is multiplied by 3 to get the next term.
Understanding the concept of GP is essential for solving problems involving sequences, series, and sums. Many board exam questions test the ability to:
Mastering these concepts helps in answering questions efficiently and accurately.
Find the 6th term of a GP where the first term \( a = 3 \) and the common ratio \( r = 2 \).
Solution:
Using the formula for the \( n \)th term:
\( T_n = a r^{n-1} \)
Substitute \( a = 3 \), \( r = 2 \), and \( n = 6 \):
\( T_6 = 3 \times 2^{6-1} = 3 \times 2^5 = 3 \times 32 = 96 \)
Answer: The 6th term is 96.
Find the sum of the first 8 terms of a GP with \( a = 5 \) and \( r = \frac{1}{3} \).
Solution:
Sum of first \( n \) terms is:
\( S_n = a \frac{1 - r^n}{1 - r} \)
Substitute \( a = 5 \), \( r = \frac{1}{3} \), and \( n = 8 \):
\( S_8 = 5 \times \frac{1 - \left(\frac{1}{3}\right)^8}{1 - \frac{1}{3}} = 5 \times \frac{1 - \frac{1}{6561}}{\frac{2}{3}} \)
Calculate numerator:
\( 1 - \frac{1}{6561} = \frac{6560}{6561} \)
Calculate denominator:
\( \frac{2}{3} \)
So,
\( S_8 = 5 \times \frac{6560}{6561} \times \frac{3}{2} = \frac{15}{2} \times \frac{6560}{6561} \approx 7.5 \times 0.999847 = 7.4989 \)
Answer: The sum of the first 8 terms is approximately 7.499.
Find the sum to infinity of a GP whose first term is 12 and the common ratio is \( \frac{1}{4} \).
Solution:
Since \( |r| = \frac{1}{4} < 1 \), sum to infinity exists and is:
\( S_\infty = \frac{a}{1 - r} \)
Substitute \( a = 12 \), \( r = \frac{1}{4} \):
\( S_\infty = \frac{12}{1 - \frac{1}{4}} = \frac{12}{\frac{3}{4}} = 12 \times \frac{4}{3} = 16 \)
Answer: The sum to infinity is 16.
Find the 5th term of a GP if the 3rd term is 18 and the 6th term is 486.
Solution:
Let the first term be \( a \) and common ratio be \( r \).
Given:
\( T_3 = a r^{2} = 18 \)
\( T_6 = a r^{5} = 486 \)
Divide \( T_6 \) by \( T_3 \):
\( \frac{T_6}{T_3} = \frac{a r^{5}}{a r^{2}} = r^{3} = \frac{486}{18} = 27 \)
So,
\( r^{3} = 27 \Rightarrow r = 3 \)
Find \( a \):
\( a r^{2} = 18 \Rightarrow a \times 3^{2} = 18 \Rightarrow a \times 9 = 18 \Rightarrow a = 2 \)
Find \( T_5 \):
\( T_5 = a r^{4} = 2 \times 3^{4} = 2 \times 81 = 162 \)
Answer: The 5th term is 162.
The sum of the first 4 terms of a GP is 30 and the sum of the next 4 terms is 120. Find the first term and common ratio.
Solution:
Let the first term be \( a \) and common ratio be \( r \).
Sum of first 4 terms:
\( S_4 = a \frac{1 - r^{4}}{1 - r} = 30 \)
Sum of next 4 terms (terms 5 to 8):
\( S_{5-8} = S_8 - S_4 = a \frac{1 - r^{8}}{1 - r} - a \frac{1 - r^{4}}{1 - r} = a \frac{r^{4} - r^{8}}{1 - r} = 120 \)
Divide the two sums:
\( \frac{S_{5-8}}{S_4} = \frac{a \frac{r^{4} - r^{8}}{1 - r}}{a \frac{1 - r^{4}}{1 - r}} = \frac{r^{4} (1 - r^{4})}{1 - r^{4}} = r^{4} = \frac{120}{30} = 4 \)
So,
\( r^{4} = 4 \Rightarrow r = \sqrt[4]{4} = \sqrt{2} \)
Now, substitute \( r = \sqrt{2} \) into the first sum equation:
\( 30 = a \frac{1 - (\sqrt{2})^{4}}{1 - \sqrt{2}} = a \frac{1 - 4}{1 - \sqrt{2}} = a \frac{-3}{1 - \sqrt{2}} \)
Calculate denominator:
\( 1 - \sqrt{2} \approx 1 - 1.414 = -0.414 \)
So,
\( 30 = a \times \frac{-3}{-0.414} = a \times 7.246 \Rightarrow a = \frac{30}{7.246} \approx 4.14 \)
Answer: First term \( a \approx 4.14 \), common ratio \( r = \sqrt{2} \).
| Concept | Formula | Conditions |
|---|---|---|
| General term (nth term) of GP | \( T_n = a r^{n-1} \) | \( a \) = first term, \( r \neq 0 \) |
| Sum of first n terms | \( S_n = a \frac{1 - r^n}{1 - r} \) | \( r \neq 1 \) |
| Sum of first n terms (if \( r = 1 \)) | \( S_n = n \times a \) | All terms equal |
| Sum to infinity | \( S_\infty = \frac{a}{1 - r} \) | \( |r| < 1 \) |
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