Full NCERT notes on arithmetic progressions, geometric sequences, nth term, sum formulas, and practice problems for Class 9 Maths Chapter 8.
A sequence is an ordered list of numbers where each number is called a term. We denote the terms as \( t_1, t_2, t_3, \dots, t_n \), where the subscript \( n \) indicates the position of the term.
Notice that every square number is the sum of consecutive odd numbers:
• \( 1 = 1 \)
• \( 4 = 1 + 3 \)
• \( 9 = 1 + 3 + 5 \)
• \( 16 = 1 + 3 + 5 + 7 \)
Calculates the term directly from its position number \( n \), without knowing previous terms! (e.g., \( u_n = 2n - 1 \) for odd numbers).
Calculates a term using the value of earlier terms in the sequence! (e.g., \(t_1 = 1, t_n = t_{n-1} + 3\)).
Problem: A sequence is generated by the explicit formula \( s_n = 5n - 2 \).
(i) Find the 62nd term.
(ii) Is 471 a term of this sequence?
Solution:
(i) Substitute \(n = 62\): \(s_{62} = 5(62) - 2 = 310 - 2 = \mathbf{308}\).
(ii) Set \( 5n - 2 = 471 \implies 5n = 473 \implies n = 4735 = 94.6 \).
Since position \( n \) must be a natural number (\( 1, 2, 3, \dots \)), 471 is NOT a term of this sequence!
An Arithmetic Progression (AP) is a sequence in which the difference between any two consecutive terms is a constant value called the common difference (\( d \)):
When the ordered pairs \( (n, t_n) \) of an AP are plotted on a Cartesian coordinate plane, all points lie strictly on a STRAIGHT LINE!
Problem: A taxi service charges a fixed booking fee of ₹200 plus ₹40 per kilometre travelled.
(i) Write the AP sequence for total fare for 1 km, 2 km, 3 km.
(ii) Calculate the total fare for a 10 km trip.
Solution:
• 1 km fare: \( t_1 = 200 + 40(1) = 240 \)
• 2 km fare: \( t_2 = 200 + 40(2) = 280 \)
• 3 km fare: \( t_3 = 200 + 40(3) = 320 \)
Sequence: \( 240, 280, 320, \dots \) (AP with \( a = 240, d = 40 \)).
\( t_n = 240 + (n - 1)40 = 200 + 40n \).
For \(n = 10\text{ km}\): \(t_{10} = 200 + 40(10) = 200 + 400 = 600\text{ rupees}\).
To find the sum \( S_n = 1 + 2 + 3 + \dots + n \), we write the sum forward and backward and add them pair-by-pair:
First recorded in Āryabhaṭa’s Āryabhaṭīya (499 CE, Chapter 2, Verse 19), stating that the sum equals the average of first and last terms multiplied by number of terms!
Problem: Calculate the sum of consecutive numbers \( 25 + 26 + 27 + \dots + 58 \).
Solution:
\[25 + 26 + \dots + 58 = S_{58} - S_{24}\]
\[S_{58} = \frac{58 \times 59}{2} = 29 \times 59 = 1711\]
\[S_{24} = \frac{24 \times 25}{2} = 12 \times 25 = 300\]
\[ Sum = 1711 - 300 = \mathbf{1411} \]
A Geometric Progression (GP) is a sequence in which each term after the first is obtained by multiplying the previous term by a constant multiplier called the common ratio (\( r \)):
Unlike an AP (which forms a straight line), when points \( (n, t_n) \) of a GP are plotted on a graph, they form an exponential curve!
A fractal is a self-similar shape that repeats at different scales. The Sierpiński Triangle (constructed by repeatedly removing central equilateral triangles) illustrates two simultaneous GPs:
| Stage (\( n \)) | Stage 0 | Stage 1 | Stage 2 | Stage 3 | Stage \( n \) Rule |
|---|---|---|---|---|---|
| Black Triangles (\( t_n \)) | \( 1 = 3^0 \) | \( 3 = 3^1 \) | \( 9 = 3^2 \) | \( 27 = 3^3 \) | \( \mathbf{t_n = 3^n} \) (Growing GP, \( r = 3 \)) |
| Shaded Area (\( s_n \)) | 1 | \( 34 \) | \( (34)^2 \) | \( (34)^3 \) | \( \mathbf{s_n = (34)^n} \) (Shrinking GP, \( r = 34 \)) |
Problem: A ball is dropped from a height of 24 feet. Each time it bounces back to \( 34 \) (75%) of its previous height.
Write the height sequence for 3 bounces and check after which bounce it falls below \( 16 \) of original height (4 feet).
Solution:
• Initial Height: 24 ft
• Bounce 1: \( 24 × 0.75 = 18 ft \)
• Bounce 2: \( 18 × 0.75 = 13.5 ft \)
• Bounce 3: \( 13.5 × 0.75 = 10.125 ft \)
• Bounce 4: \( 7.594 ft \), Bounce 5: \( 5.695 ft \), Bounce 6: \( 4.271 ft \), Bounce 7: \( \mathbf{3.203 ft} \).
After the 7th bounce, the height (3.20 ft) falls below 4 ft (\( 16 \) of 24 ft)!