geometric progression

After learning arithmetic progression let us learn geometric progression, which is also known as a geometric sequence.
Geometric progression(G.P) is a sequence in which the ratio of any term to its predecessors is always the same number, which is constant. The ratio is called the common ratio. If 'a' denotes first term and 'r' the common ratio in a geometric progression then the standard geometric progression is a, ar, ar², …
Examples:
(i) 1, 3, 9, 27, 81, …
common ratio \(\frac{3}{1} = \frac{9}{3} = \frac{27}{9} = \frac{81}{27} =… = 3\)
(ii)  \(1, -\frac{1}{2}, \frac{1}{4}, -\frac{1}{8}, …\)
Common Ratio \(\frac{-1/2}{1} =\frac{1/4}{-1/2} = \frac{-1/8}{1/4} = … = \frac{-1}{2}\)

The n^th Term of a Geometric Progression

If a is the first term, common ratio is r and number of term is n then
t₁(first term) = a⋅r¹⁻¹ = a
t₂(second term) = a⋅r²⁻¹ = ar
t₃(third term) = a⋅r³⁻¹ = ar²
Therefore,

nth term of geometric progression, tn = arn−1

Question 1: Determine if the sequence 0.2, 0.02, 0.002, 0.0002, ... is geometric or not.
Solution: Divide the terms by their predecessor and check whether a common ratio exists.
.02 ÷ 0.2 = 0.1, .002 ÷ 0.02 = 0.1, .0002 ÷ 0.002 = .1
As they have common ratio = 0.1, therefore this sequence is a geometric progression.

Question 2: Find the 6^th term of geometric progression 3, 15, 75, 375, ...
Solution: t₆ = 3 ⋅ 5⁶⁻¹ = 3 ⋅ 5⁵ = 3 ⋅ 3125 = 9375

Question 3: n^th term of geometric progression is 3⋅2ⁿ⁻¹, find the first and second term.
Solution: t₁ = 3⋅2⁰ = 3, t₂ = 3⋅2²⁻¹ = 6

Sum of n Terms of a Geometric Progression

If a is the first term, r is the common ratio and S_n is the sum of n terms of the geometric progression, then
S_n = a + ar + ar² + ar³ + … + arⁿ⁻¹

\(S_n = \frac {a(r^n - 1)}{r - 1}\), when r > 1   \(S_n = \frac {a(1 - r^n)}{1 - r}\), when r < 1

Question 1: Find the sum of the geometric progression, 3, -6, 12, -24, 48, ... to 10 terms
Solution: Common ration = -6/3 = -2
as r < 1, therefore \(S_n = \frac {a(1 - r^n)}{1 - r}\)
\(S_1 = \frac { 3(1-(-2)^{10})} { 1 - (-2)} \\ S_1 = \frac{3(1-1024)}{3} \\ S_1 = -1023\)

Geometric Mean (G.M)

The geometric mean of two positive numbers a and b is the number \(\sqrt {ab}\). Therefore the geometric mean of 8 and 32 is \(\sqrt {8.32}\) = \(\sqrt {256} = 16\)

Did you notice that the three numbers 8, 16, and 32 are three consecutive terms of geometric progression.  So the geometric mean gives us a way of finding a value in between widely different values. To find the geometric mean of n numbers, multiply all n numbers and take the n^th root. Therefore,
Geometric Mean of n numbers a₁ to a_n is : \(\sqrt[n]{a_1\times a_2 \times a_3 \times…\times a_n}\)

The geometric mean of 3, 27 is 9, which means the area of a rectangle of sides 3 and 27 is the same as the area of a square of side 9.
Question: Find the geometric mean of 4, 10, 12, 20, 24
Solution: Geometric Mean = \(\sqrt[5]{4\times10\times 12\times20\times24}\)
Geometric Mean = 11.816