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Suppose W1 = 1, W2 = 2 and for n > 2, Wn = W1 + W2 + ··· + Wn-2 + 2. Find the values of W1, W2, …, W8. Do you recognise this sequence?
Given W1 = 1, W2 = 2 and Wn = W1 + ... + Wn-2 + 2: W3 = W1 + 2 = 1 + 2 = 3 W4 = (1 + 2) + 2 = 5 W5 = (1 + 2 + 3) + 2 = 8 W6 = 8 + 5 = 13 W7 = 13 + 8 = 21 W8 = 21 + 13 = 34. The first eight values are: 1, 2, 3, 5, 8, 13, 21, 34. Notice that Wn - Wn-1 = Wn-2, which gives Wn = Wn-1 + Wn-2. This is theRead more
Given W1 = 1, W2 = 2 and Wn = W1 + … + Wn-2 + 2:
W3 = W1 + 2 = 1 + 2 = 3
W4 = (1 + 2) + 2 = 5
W5 = (1 + 2 + 3) + 2 = 8
W6 = 8 + 5 = 13
W7 = 13 + 8 = 21
W8 = 21 + 13 = 34.
The first eight values are: 1, 2, 3, 5, 8, 13, 21, 34.
Notice that Wn – Wn-1 = Wn-2, which gives Wn = Wn-1 + Wn-2.
This is the Virahanka-Fibonacci sequence.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/
See lessSuppose P1 = 1, P2 = 2 and for n > 2, Pn = P1 + P2 + ··· + Pn-1 + 1. Find the values of P1, P2, …, P8. Can you find a simpler recursive formula for Pn? Can you give an explicit formula?
Given P1 = 1, P2 = 2 and Pn = P1 + P2 + ... + Pn-1 + 1: P3 = 1 + 2 + 1 = 4 P4 = 4 + 4 = 8 P5 = 8 + 8 = 16 P6 = 32, P7 = 64, P8 = 128. Values are: 1, 2, 4, 8, 16, 32, 64, 128. Simpler recursive rule: Pn = (P1 + ... + Pn-2 + 1) + Pn-1 = Pn-1 + Pn-1 = 2Pn-1 for n >= 3 (with P1 = 1, P2 = 2). ExplicitRead more
Given P1 = 1, P2 = 2 and Pn = P1 + P2 + … + Pn-1 + 1:
P3 = 1 + 2 + 1 = 4
P4 = 4 + 4 = 8
P5 = 8 + 8 = 16
P6 = 32, P7 = 64, P8 = 128.
Values are: 1, 2, 4, 8, 16, 32, 64, 128.
Simpler recursive rule:
Pn = (P1 + … + Pn-2 + 1) + Pn-1 = Pn-1 + Pn-1 = 2Pn-1 for n >= 3 (with P1 = 1, P2 = 2).
Explicit formula:
P1 = 1 and Pn = 2ⁿ⁻¹ for all n >= 2.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/
See lessThe sum of the first three terms of a geometric progression is 26 and the sum of their squares is 364. Find the terms of the GP.
Let the terms be a, ar and ar². Given: a(1 + r + r²) = 26 Squaring: a²(1 + r + r²)² = 676 Sum of squares: a²(1 + r² + r⁴) = 364 Using identity (1 + r² + r⁴) = (1 + r + r²)(1 - r + r²): Dividing gives (1 + r + r²) / (1 - r + r²) = 676 / 364 = 13 / 7. 7 + 7r + 7r² = 13 - 13r + 13r² 6r² - 20r + 6 = 0,Read more
Let the terms be a, ar and ar².
Given:
a(1 + r + r²) = 26
Squaring: a²(1 + r + r²)² = 676
Sum of squares:
a²(1 + r² + r⁴) = 364
Using identity (1 + r² + r⁴) = (1 + r + r²)(1 – r + r²):
Dividing gives (1 + r + r²) / (1 – r + r²) = 676 / 364 = 13 / 7.
7 + 7r + 7r² = 13 – 13r + 13r²
6r² – 20r + 6 = 0, so 3r² – 10r + 3 = 0.
Thus r = 3 or r = 1/3.
If r = 3, a = 2, terms are 2, 6, 18.
If r = 1/3, a = 18, terms are 18, 6, 2.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/
See lessIf the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.
Let the GP have first term a and common ratio r. Given: 4th term: x = ar³ 10th term: y = ar⁹ 16th term: z = ar¹⁵ Now check the square of the middle term: y² = (ar⁹)² = a²r¹⁸. Now calculate the product of the first and third terms: x x z = (ar³) x (ar¹⁵) = a² x r³⁺¹⁵ = a²r¹⁸. Since y² = x x z, the raRead more
Let the GP have first term a and common ratio r.
Given:
4th term: x = ar³
10th term: y = ar⁹
16th term: z = ar¹⁵
Now check the square of the middle term:
y² = (ar⁹)² = a²r¹⁸.
Now calculate the product of the first and third terms:
x x z = (ar³) x (ar¹⁵) = a² x r³⁺¹⁵ = a²r¹⁸.
Since y² = x x z, the ratio y / x equals z / y = r⁶.
Therefore, x, y and z are in GP.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/
See lessThe sum of the first three terms of a GP is 13/12 and their product is –1. Find the common ratio and the terms.
Let the three terms be a/r, a and ar. Their product is: (a/r) x a x ar = -1 a³ = -1, which gives a = -1. Their sum is 13/12: -1/r - 1 - r = 13/12 -(1/r + r) = 13/12 + 1 = 25/12 (r² + 1) / r = -25/12 12r² + 25r + 12 = 0 (3r + 4)(4r + 3) = 0. So r = -4/3 or r = -3/4. When r = -4/3, terms are: 3/4, -1,Read more
Let the three terms be a/r, a and ar.
Their product is:
(a/r) x a x ar = -1
a³ = -1, which gives a = -1.
Their sum is 13/12:
-1/r – 1 – r = 13/12
-(1/r + r) = 13/12 + 1 = 25/12
(r² + 1) / r = -25/12
12r² + 25r + 12 = 0
(3r + 4)(4r + 3) = 0.
So r = -4/3 or r = -3/4.
When r = -4/3, terms are: 3/4, -1, 4/3.
When r = -3/4, terms are: 4/3, -1, 3/4.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/
See less