Geometric progression (GP) is a crucial mathematical concept that is used in various real-world scenarios, such as calculating compound interest, population growth, and even physics problems like the distance traveled by a bouncing ball. This article will cover the basics of geometric progression, its formulas, and practical examples to solidify your understanding.
A Geometric Progression or Geometric Sequence is a sequence of numbers in which each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio.
For example:
1. General Form of GP: A geometric progression with the first term aa and common ratio rr is represented as: a,ar,ar2,ar3,…a, ar, ar^2, ar^3, \dots
2. nth Term of GP: The nth term TnT_n of a geometric sequence is given by:
Tn=a⋅rn−1T_n = a \cdot r^{n-1}
Where:
3. Sum of First n Terms of GP: The sum of the first nn terms SnS_n is:
Sn=a⋅1−rn1−r, for r≠1S_n = a \cdot \frac{1 – r^n}{1 – r}, \text{ for } r \neq 1
4. Sum of Infinite Terms (for ∣r∣<1|r| < 1): The sum of an infinite geometric series is:
S∞=a1−r, where ∣r∣<1S_\infty = \frac{a}{1 – r}, \text{ where } |r| < 1
Problem: The number of bacteria in a culture doubles every hour. Initially, there are 50 bacteria. How many bacteria are present after 12 hours?
Solution: The bacteria growth forms a geometric progression:
Using the nth term formula:
T12=50⋅212−1=50⋅2048=102,400 bacteriaT_{12} = 50 \cdot 2^{12-1} = 50 \cdot 2048 = 102,400 \text{ bacteria}
Problem: A ball is dropped from a height of 128 meters. After each bounce, it rises to half the height it fell from before. What is the total distance traveled by the ball?
Solution: The total distance traveled forms a geometric series:
The total distance is given by:
Total Distance=128+1281−1/2=128+128×2=384 meters\text{Total Distance} = 128 + \frac{128}{1 – 1/2} = 128 + 128 \times 2 = 384 \text{ meters}
Problem: Insert three geometric means between 2 and 818\frac{81}{8}.
Solution: Let the three geometric means be x,y,zx, y, z. The sequence becomes: 2,x,y,z,8182, x, y, z, \frac{81}{8}.
The common ratio rr can be calculated by:
r4=81/82 ⟹ r=32r^4 = \frac{81/8}{2} \implies r = \frac{3}{2}
Thus, the geometric means are:
The geometric means are 3, 92\frac{9}{2}, and 274\frac{27}{4}.
Solution: Let aa be the first term, and rr be the common ratio. From the given conditions:
S∞=a1−r=48S_\infty = \frac{a}{1 – r} = 48 a+ar=36a + ar = 36
Solving these equations gives:
Solution: This is a challenging sum that involves a more complex formula for the terms in the series:
Sn=129((10n−1))−12n9S_n = \frac{12}{9} \left( (10^n – 1) \right) – \frac{12n}{9}Sn=912((10n−1))−912n
Geometric Progression (GP) is a fundamental concept that is widely used in various fields, including biology, physics, and finance. By mastering the formulas and understanding the behavior of sequences, you can solve real-world problems effectively. With practice, you’ll be able to identify geometric sequences in various scenarios and apply the appropriate formulas to find solutions.