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Ex 8.3, 2
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
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Exercise Set 8.3
- End-of-Chapter Exercises
Transcript
Ex 8.3, 2 Find the 10th and nth terms of the GP: 5, 25, 125,β¦. Given GP 5, 25, 125, β¦. Here, First term = a = 5 Common ratio = r = 25/5 = 5 We know that for a GP nth term = π_π=γππγ^(πβπ) Finding nth term Now, π_π=γππγ^(πβπ) Putting a = 5, r = 5 =π Γ π^(πβπ) =5^1 Γ 5^(πβ1) =5^(1 + π β 1) =π^π Thus, nth term of GP is π^π Finding 10th term For 10th term, we put n = 10 in nth term Now, π_π=γππγ^(πβπ) Putting a = 5, r = 5 =π Γ π^(πβπ) =5^1 Γ 5^(πβ1) =5^(1 + π β 1) =π^π π_π=5^π Putting n = 10 π_ππ=π^ππ Hence, 10th term of GP is π^ππ
