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Ex 8.3, 6
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
- Virahānka–Fibonacci 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, 6 Which term of the sequence 2, 2√2,4,… is 128? Given GP 2, 2√2,4,… Here, First term = a = 2 Common ratio = r = (2√2)/2 = √𝟐 We need to find which term is 128 So, an = 128 We need to find n We know that for a GP 𝒂_𝒏=〖𝒂𝒓〗^(𝒏−𝟏) Putting a = 2, r = √2, an = 128 𝟏𝟐𝟖=𝟐 × (√𝟐 )^(𝒏−𝟏) 128/2=(√(2 ))^(𝑛−1) 64=(√2 )^(𝑛−1) 64=(√2 )^(𝑛−1) 64=(2^(1/2) )^(𝑛 − 1) 64=(2)^((𝑛 − 1)/2) 𝟐^𝟔=(𝟐)^((𝒏 − 𝟏)/𝟐) Comparing powers Comparing powers 𝟔=(𝒏 − 𝟏)/𝟐 6 × 2=𝑛−1 12=𝑛−1 12+1=𝑛 13=𝑛 𝒏=𝟏𝟑 Thus, it is the 13th term
