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![[Class 9] Which term of the sequence tn = 5n – 3 for n ≥ 1 is 607? - Exercise Set 8.1](https://cdn.teachoo.com/1663059c-4f0f-4ce0-b2b5-8f51b47db4eb/slide65.jpg)
Ex 8.1, 4 - Chapter 8 - Predicting What Comes Next: Exploring Sequences & Progress
Last updated at August 12, 2026 by Teachoo
Maths 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
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Transcript
Ex 8.1, 4 Which term of the sequence π‘_π=5πβ3 for πβ₯1 is 607? To check which term is 607 of the sequence We put π‘_π = 607 and find n Given π_π=ππβπ Putting π_π = 607 607 = 5n β 3 607 + 3 = 5n 610 = 5n 5n = 610 n = 610/5 n = 122 Thus, 607 is the 122nd term of the sequence
