Exploring Some More Progressions Class 9 Notes Advanced Maths Chapter 6

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Exploring Some More Progressions Notes Class 9 Advanced Maths

Introduction

You already know how to add numbers in a straight line — an AP But the real world rarely grows in straight lines. Money compounds, populations multiply, rubber balls bounce and paper folds, all of these follow a Geometric Progression.

This chapter explains how to determine the total of any GP in one formula, saving you from adding hundreds of terms one by one. More fascinatingly, it shows you that even an infinite number of terms can have a finite sum — something that feels impossible until you see why a bouncing ball eventually stops at a predictable total distance.

The chapter also introduces the Method of Differences — a powerful technique for sequences that are neither AP nor GP turning what looks like a random pattern into something completely solvable.

Together, these three ideas — finite GP infinite GP and Method of Differences — form a toolkit that goes far beyond the classroom, connecting to real-life problems, ancient Indian mathematics (Madhava of the Kerala School) and the foundations of higher mathematics you will encounter in later years.

Sum of the First n Terms of a Geometric Progression

A Geometric Progression (GP) is a sequence of the form a, ar, ar², ar³,… where a is the first term and r(non-zero) is the common ratio. The n^th term is given by t_n = ar^n-1.

Fundamentals:

1. The ratio of any term to its previous term is always equal to r (constant).
2. Both a and r must be non-zero in a GP
3. When r = 1, all terms are equal and S_n = na.
4. The formula for S_n is derived by multiplying the series by r and subtracting from the original.

Points to remember:

1. Use the form \(\frac{a\left(1-r^n\right)}{1-r}\) when |r| < 1 to avoid sign errors. • Use the form \(\frac{a\left(r^n-1\right)}{r-1}\) when |r| > 1 for convenience.
2. Always state the condition r ≠ 1 when writing the formula.
3. For series like 7 + 77 + 777 + …, rewrite each term as \(\frac{7}{9}\) × (10ⁿ – 1) and split into a GP part and a separate -n correction.

Formulae:
Sum of the First n Terms:

1. S_n = \(\frac{a\left(1-r^n\right)}{1-r}\), r ≠ 1
2. S_n = na, when r = 1

Example 1:
Riya saves ₹500 in January. Each month she doubles her savings. So, her monthly savings are: ₹500, ₹1000, ₹2000, ₹4000, …. How much does she save in total over 6 months?
Solution:
Given: First term a = ₹500, common ratio r = 2, number of terms n = 6
Each month her savings double, so the ratio between consecutive terms is constantly 2. This confirms a GP with a = 500 and r = 2.
Since r > 1, we use:
S_n = \(\frac{a\left(r^n-1\right)}{r-1}\)
S₆ = \(\left(\frac{2^6-1}{2-1}\right)\)
= 500 × \(\left(\frac{64-1}{1}\right)\)
= 500 × 63 = ₹31,500
Riya saves a total of ₹31,500 over 6 months.

Example 2:
A radioactive substance has an initial mass of 800 g. Every year it reduces to half its mass. What is the total mass lost by the substance over 5 years?
Solution:
Mass each year: 400, 200, 100, — a GP with
a = 400, r = \(\frac{1}{2}\), n = 5′
Since, |r| = \(\frac{1}{2}\) < 1, use S_n = \(\frac{a\left(1-r^n\right)}{1-r}\)
S₅ = \(400 \times \frac{1-\left(\frac{1}{2}\right)^5}{1-\frac{1}{2}}=400 \times \frac{\left(1-\frac{1}{32}\right)}{\left(\frac{1}{2}\right)}\)
= \(400 \times\left(\frac{31}{32}\right) \times 2\)
= 25 × 31 = 775 g
∴ Total mass lost over 5 years = 775 g

Infinite Geometric Progression

When the absolute value of the common ratio is less than 1, the terms of a GP keep shrinking and approach zero. This allows the sum of infinitely many terms to converge to a definite finite value.

Fundamentals:

1. If | r | < 1, then rⁿ —» 0 as n → ∞, making the infinite sum finite.
2. If | r | ≥ 1, the terms do not shrink and the infinite sum does not exist.
3. The condition -1 < r < 1 must always be verified before applying the infinite sum formula.
4. Historical note: Madhava of the Kerala School used infinite series to calculate n correct to 11 decimal places.

Points to remember:

1. Always check the convergence condition -1 < r < 1 first – never skip this step.
2. If the condition fails, do not apply the formula; state the convergence condition to that the sum does not exist.
3. For an infinite GP with first term x and sum 4: the possible values of x satisfy 0 < x < 8 (derived by applying the convergence condition to
   r = 1 – \(\frac{x}{4}\))

Example 1:
Sum of Infinite Terms: (only when -1 < r < 1):
S_∞ = \(\frac{a}{1-r}\)
A rubber ball is dropped from a height of 18 m. Each time it hits the ground, it bounces back to \(\frac{2}{3}\) of the height it fell from. Find the total distance the ball travels before coming to rest.
Solution:
Rebound heights: 12, 8, \(\frac{16}{3}\),… — a GP with a = 12, r = \(\frac{2}{3}\). Since | r | < 1, infinite sum exists.
∵ r = \(\frac{2}{3}\), since -1 < \(\frac{2}{3}\) < 1.
Apply S_∞ = \(\frac{a}{1-r}\)
Sum of all rebound heights (upward journeys only) is given by-
S_∞ = \(\frac{12}{\left(1-\frac{2}{3}\right)}=\frac{12}{\frac{1}{3}}\) = 36 m
Ball travels down 18 m, then up+down for each bounce = 2 × S_∞.
Total distance = 18 + 2 × 36 = 18 + 72 = 90 m
∴ Total distance travelled by the ball = 90 m

Example 2:
An artist draws a square of side 24 cm. He joins the midpoints of its sides to form a smaller square inside and repeats this process infinitely. Find the sum of the perimeters of all the squares formed.
Solution:
Each new square’s side = \(\left(\frac{1}{\sqrt{2}}\right)\) × previous side.
Perimeters: 96, \(\frac{96}{\sqrt{2}}, \frac{96}{2}\), … -GP with a = 96, r = \(\frac{1}{\sqrt{2}}\), we know that
When midpoints of a square of side s are joined, new side = \(\frac{s}{\sqrt{2}}\).
So, r = \(\frac{1}{\sqrt{2}}=\frac{1}{1.4142}\) ≈ 0.707. Since |r| < 1.
S_∞ = \(\frac{a}{1-r}=\frac{96}{\left(1-\frac{1}{\sqrt{2}}\right)}\)
Multiply numerator and denominator by \(\left(1+\frac{1}{\sqrt{2}}\right)\)
S_∞ = \(\frac{96\left(1+\frac{1}{\sqrt{2}}\right)}{1-\frac{1}{2}}=\frac{96\left(1+\frac{1}{\sqrt{2}}\right)}{\frac{1}{2}}\)
= \(192\left(1+\frac{1}{\sqrt{2}}\right)\) ≈ 192 × 1.707 ≈ 327.8 cm.
∴ Sum of all perimeters = \(192\left(1+\frac{1}{\sqrt{2}}\right)\) ≈ 327.8 cm.

Method of Differences and Combinatorics

For sequences that are neither AP nor GR if the differences between consecutive terms form an AP (or can be reduced to one), the n^th term and sum can be found using the Method of Differences combined with combinatorial expressions.

Fundamentals:
Write the sequence in Row
1. Compute successive differences to get Row 2, Row 3, etc.

• If Row 2 is an AP (i.e., Row 3 is constant), the formula for tn uses three parameters: b, a, d.
• b = first term of original sequence (Row 1)
• a = first term of Row 2 (1^st differences)
• d = first term of Row 3 (2^nd differences)
• If Row 3 is not constant, introduce d₁ (first term of Row 4) and extend the formula accordingly.
• This concept is rooted in the ancient Indian method Varasankalita discussed by Narayana in Ganita Kaumudi.

Points to remember:

1. Always write all rows of differences before identifying b, a, d.
2. Do not apply the formula until Row 2 is confirmed to be an AP
3. If Row 3 has 4 or more rows of differences, extend the formula with additional C(n -1, 3), C(n -1, 4) terms.
4. The combinatorics method is a shortcut; the subtraction method (writing S twice and subtracting) gives the same result.

Formula:
n^th term (when Row 2 is an AP):
t_n = b + a(n – 1) + d.\(\frac{(n-1)(n-2)}{2}\)
Equivalently using Combinations:
t_n = b.C(n – 1, 0) + a . C(n – 1, 1) + d . C(n – 1, 2)
Sum of the first n terms:
S_n = bn + \(a \cdot \frac{n(n-1)}{2}+d \cdot \frac{n(n-1)(n-2)}{3 \times 2}\)
or S_n = b.C(n, 1) + a.C(n, 2) + d.C(n, 3)
Combination Formula (for reference):
C(n, r) = \(\frac{n!}{[r!\times(n-r)!]}\)

Example 1:
A builder stacks bricks in rows. The number of bricks in successive rows follows the pattern: 5, 11, 19, 29, 41, … Find the number of bricks in the 10^th row and the total bricks used in 10 rows.
Solution:
The differences between terms are 6, 8,10,12 – an AP So, Method of Differences applies.
Write difference rows:
Row 1: 5, 11, 19, 29, 41 → b = 5
Row 2: 6, 8, 10, 12 → a = 6 (AP)
Row 3: 2, 2, 2 → d = 2
Apply n^th term formula with n = 10:
t_n = b + a(n -1) + \(d \cdot \frac{(n-1)(n-2)}{2}\)
t₁₀ = 5 + 6(9) + \(2 \cdot \frac{(9)(8)}{2}\)
= 5 + 54 + 72 = 131
Apply sum formula for n = 10:
S_n = b_n + \(a \cdot \frac{n(n-1)}{2}+d \cdot \frac{n(n-1)(n-2)}{3 \times 2}\)
= 5(10) + \(6 \cdot \frac{10 \times 9}{2}+2 \cdot \frac{10 \times 9 \times 8}{6}\)
= 50 + 270 + 240 = 560
As the 10^th row has 131 bricks.
Total in 10 rows = 560 bricks

Example 2:
In a cricket tournament, the number of runs scored in i successive overs by a team follows the pattern: 1, 4, 10, 20, 35, … Find the runs scored in the 8^th over and the total runs after 8 overs.
Solution:
Row 2 differences: 3, 6, 10, 15 — not constant. Row 3: 3, 4, 5 — still not constant. Row 4: 1, 1 — constant.
Extended method needed,
Write all difference rows:
Row 1: 1, 4, 10, 20, 35 → b = 1
Row 2: 3, 6, 10,15 → a = 3
Row 3: 3, 4, 5 → d = 3
Row 4: 1, 1 → d₁ = 1 (constant)
Extended n^th term formula (4 rows):
t_n = b.C(n – 1, 0) + a.C(n – 1, 1) + d.C(n – 1, 2) + d₁.C(n – 1, 3)
For n = 8, compute each C(7, r):
C(7, 0) = 1, C(7, 1) = 7, C(7, 2) = 21, C(7, 3) = 35
Now calculate each combination using the formula
C(n, r ) = \(\frac{n!}{[r!\times(n-r)!]}\)
C(7, 0) = \(\frac{7!}{[0!\times 7!]}\) = 1
C(7, 1) = \(\frac{7!}{[1!\times 6!]}=\frac{7}{1}\) = 7
C(7, 2) = \(\frac{7!}{[2!\times 5!]}=\frac{7 \times 6}{2 \times 1}=\frac{42}{2}\) = 21
C(7, 3) = \(\frac{7!}{[3!\times 4!]}=\frac{7 \times 6 \times 5}{3 \times 2 \times 1}\)
= \(\frac{210}{6}\) = 35
t₈ = 1(1) + 3(7) + 3(21) + 1(35)
= 1 + 21 + 63 + 35
= 120
Sum S₈ = S_n = b.C(n, 1) + a.C(n, 2) + d.C(n, 3) + d₁.C(n, 4)
C(8, 1) = \(\frac{8!}{[1!\times 7!]}=\frac{8}{1}\) = 8
C(8, 2) = \(\frac{8!}{2!\times 6!}=\frac{8 \times 7}{2 \times 1}=\frac{56}{2}\) = 28
C(8, 3) = \(\frac{8!}{[3!\times 5!]}=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}\)
= \(\frac{336}{6}\) = 56
C(8, 4) = \(\frac{8!}{[4!\times 4!]}=\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}\)
= \(\frac{1680}{24}\) = 70
So, S₈ = 1.C(8, 1) + 3.C(8, 2) + 3.C(8, 3) + 1.C(8, 4)
= 1(8) + 3(28) + 3(56) + 1(70)
= 8 + 84 + 168 + 70 = 330
∵ Runs in 8^th over = 120
Total runs after 8 overs = 330