The document provides an introduction to geometric progressions (GP), explaining their properties, formulas for the n-th term and the sum of the first n terms, and applications in various fields. It also includes several examples to illustrate finding specific terms and sums within GPs. Key components discussed include the general form, common ratio, and methods to solve for unknowns in GPs.

1. Geometric Progression
• An introduction to geometric sequences and
their properties.
• By : Dr. Sabyasachi Ray

2. General Form
• A geometric progression (GP) is a sequence of
numbers where each term after the first is
found by multiplying the previous term by a
constant. The general form of a GP is:
• a, ar, ar2, ar3, ...

3. Common Ratio
• The constant factor between consecutive
terms is called the common ratio (r). If the
first term is a, then the second term is ar, the
third term is ar2, and so on.

4. Finding the n-th Term
• The n-th term (Tn) of a geometric progression
can be found using the formula:
• Tn = a * r(n-1)
• where a is the first term, r is the common
ratio, and n is the term number.

5. Sum of the First n Terms
• The sum of the first n terms (Sn) of a
geometric progression is given by:
• - For r ≠ 1: Sn = a * (1 - rn) / (1 - r)
• - For r = 1: Sn = a * n

6. Applications of Geometric
Progressions
• Geometric progressions are used in various
fields such as finance (compound interest),
biology (population growth), physics
(radioactive decay), and computer science
(algorithm analysis).

7. Examples
• 1. Find the 5th term of a GP with a = 2 and r = 3.

8. Examples
• 2. Calculate the sum of the first 4 terms of a
GP with a = 5 and r = 2.

9. Example 1
• Find the 5th term of a GP where the first term
(a) is 2 and the common ratio (r) is 3.
• Solution:
• Tn = a * r^(n-1)
• T5 = 2 * 3^(5-1)
• T5 = 2 * 3^4
• T5 = 2 * 81
• T5 = 162

10. Example 2
• Calculate the sum of the first 4 terms of a GP
where the first term (a) is 5 and the common
ratio (r) is 2.
• Solution:
• Sn = a * (1 - r^n) / (1 - r)
• S4 = 5 * (1 - 2^4) / (1 - 2)
• S4 = 5 * (1 - 16) / -1
• S4 = 5 * (-15) / -1
• S4 = 75

11. Example 3
• If the 3rd term of a GP is 16 and the 6th term is 128, find the common ratio and
the first term.
• Solution:
• T3 = ar^2 = 16
• T6 = ar^5 = 128
• Divide T6 by T3:
• (ar^5) / (ar^2) = 128 / 16
• r^3 = 8
• r = 2
• Substitute r into T3:
• a * 2^2 = 16
• a * 4 = 16
• a = 4
• First term (a) = 4
• Common ratio (r) = 2

12. Example 4
• The sum of the first 5 terms of a GP is 31 and the first term
is 1. Find the common ratio.
• Solution:
• S5 = a * (1 - r^5) / (1 - r)
• 31 = 1 * (1 - r^5) / (1 - r)
• 31 = (1 - r^5) / (1 - r)
• 31 * (1 - r) = 1 - r^5
• 31 - 31r = 1 - r^5
• r^5 - 31r + 30 = 0
• Solve for r (possible values by trial and error or factoring):
• r = 2 (valid solution)