Class 9 Mathematics Chapter 8 Predicting What Comes Next?: Exploring Sequences and Progressions – Extra Questions with Answers

Extra practice questions for Class 9 Maths Chapter 8 (Predicting What Comes Next?: Exploring Sequences and Progressions), beyond the textbook. These Class 9 Mathematics Chapter 8 important questions are handy for last-minute exam practice.

Very Short Answer Questions (1 mark)

Q1. Find the common difference of the AP: 5, 9, 13, 17, …
Ans: d = 9 − 5 = 4.

Q2. Write the formula for the nth term of an AP.
Ans: a_n = a + (n−1)d.

Q3. Is the sequence 2, 4, 8, 16 an AP? Why or why not?
Ans: No, the differences (2, 4, 8) are not constant, so it is not an AP (it’s a geometric progression instead).

Q4. Find the 5th term of the AP with first term 3 and common difference 2.
Ans: a₅ = 3 + (5−1)(2) = 3 + 8 = 11.

Q5. What is another name for the terms of a sequence that follow a specific rule?
Ans: Terms of the sequence.

Short Answer Questions (2–3 marks)

Q6. Find the sum of the first 10 terms of the AP: 2, 5, 8, 11, …
Ans: a=2, d=3, n=10. S₁₀ = 10/2 × [2(2) + (10−1)(3)] = 5 × [4 + 27] = 5 × 31 = 155.

Q7. The 3rd term of an AP is 12, and the 7th term is 24. Find the first term and common difference.
Ans: a + 2d = 12 and a + 6d = 24. Subtracting: 4d = 12, so d = 3. Substituting back: a + 2(3) = 12, so a = 6.

Q8. A stack of logs has 20 logs in the bottom row, 19 in the next row, and so on, decreasing by 1 each row, up to a single log at the top. Find the total number of rows and the total number of logs, recognising this as an AP.
Ans: This is an AP with a=20, d=−1, going down to a_n=1. Using a_n=a+(n−1)d: 1 = 20 + (n−1)(−1), so (n−1) = 19, giving n = 20 rows. Total logs: S₂₀ = 20/2 × (20+1) = 10 × 21 = 210 logs.

Higher-Order Thinking / Application Questions

Q9. A person starts saving ₹500 in the first month and increases their savings by ₹100 every subsequent month. Using the AP formulas, find how much they save in the 12th month, and their total savings over the first 12 months.
Ans: a=500, d=100. 12th month savings: a₁₂ = 500 + (12−1)(100) = 500 + 1100 = ₹1600. Total over 12 months: S₁₂ = 12/2 × [2(500) + (12−1)(100)] = 6 × [1000 + 1100] = 6 × 2100 = ₹12600.

Q10. Prove, using algebra, that if three numbers a, b, c are in arithmetic progression, then 2b = a + c.
Ans: If a, b, c are in AP, the common difference between consecutive terms is equal: b − a = c − b (since both equal the common difference d). Rearranging: b + b = a + c, so 2b = a + c. This confirms that the middle term of three numbers in AP is the average of the other two.

Written by Satish

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