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Example 9 - Chapter 8 - Predicting What Comes Next: Exploring Sequences & Progress
Last updated at June 9, 2026 by Teachoo
Class 9
Chapter 8 - Predicting What Comes Next: Exploring Sequences & Progress
- Definition
- Explicit Rule for a Sequence
- Recursive Rule for a Sequence
- VirahΔnkaβFibonacci sequence
- Exercise Set 8.1
- Arithmetic Progressions
- Visualising an AP
- Sum of the First n Natural Numbers
- Exercise Set 8.2
-
Geometric Progressions
- Fractals as a GP Sequence
- Visualising a GP
- Exercise Set 8.3
- End-of-Chapter Exercises
Transcript
Example 9 Check whether the sequence 5, 15/4, 45/16, 135/64,β¦ is a geometric progression and find its nth term. Given sequence 5, 15/4, 45/16, 135/64,β¦ If its GP, then common ratio is same So, we need to show (π^ππ ππππ)/(π^ππ ππππ)=" " (π^ππ ππππ)/(π^ππ ππππ) Now, (ππ/π)/π=(ππ/ππ)/(ππ/π) 15/(4 Γ 5)=45/(16 ) Γ 4/15 3/4=3/4 Γ 1/1 π/π=π/π Since common ratio is same, it is a GP Now, we need to find nth term Here, First term = a = 5 Common ratio = r = π/π Now, nth term = π_π = ππ^(πβ1) Putting values = π Γ (π/π)^(πβπ)
