This MCQ module is based on: Sequences and Progressions — Exercises
TOPIC 25 OF 25
Sequences and Progressions — Exercises
🎓 Class 9
Mathematics
CBSE
Theory
Ch 8 — Predicting What Comes Next: Exploring Sequences and Progressions
⏱ ~40 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Sequences and Progressions — Exercises
Targeting Class 9 level in Sequences Series, with Intermediate difficulty.
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Chapter 8 Summary — Quick Reference
All formulas at a glance
- Sequence: ordered list \(a_1, a_2, a_3, \ldots\)
- AP: sequence with constant difference \(d = a_{n+1} - a_n\).
- n-th term: \(a_n = a + (n-1)d\)
- Sum to n terms: \(S_n = \dfrac{n}{2}\,[2a + (n-1)d]\) or \(\dfrac{n}{2}(a + l)\)
- Sum 1 to n: \(\dfrac{n(n+1)}{2}\)
- Sum first n odd: \(n^2\) | Sum first n even: \(n(n+1)\)
Exercise 8.1 — Sequences and the n-th Term
Q1. Identify which of the following are APs and write a, d:
(i) 3, 6, 9, 12, … (ii) −5, −1, 3, 7, … (iii) 1, 4, 9, 16, … (iv) 2, 2, 2, 2, … (v) 1, 1/2, 0, −1/2, …
(i) AP: a = 3, d = 3. (ii) AP: a = −5, d = 4. (iii) Not AP (differences 3, 5, 7 unequal). (iv) AP: a = 2, d = 0. (v) AP: a = 1, d = −1/2.
Q2. Find the indicated term:
(i) 18th term of 7, 11, 15, … (ii) 25th term of 3, 8, 13, …
(iii) 30th term of −5, −2, 1, 4, … (iv) 12th term of 100, 95, 90, …
(i) a = 7, d = 4. \(a_{18} = 7 + 17(4) = 75\). (ii) \(a_{25} = 3 + 24(5) = 123\). (iii) \(a_{30} = -5 + 29(3) = 82\). (iv) \(a_{12} = 100 + 11(-5) = 45\).
Q3. Which term of the AP 21, 18, 15, … is −81? Is any term of this AP equal to zero?
a = 21, d = −3. \(21 + (n-1)(-3) = -81 \Rightarrow (n-1)(-3) = -102 \Rightarrow n-1 = 34 \Rightarrow n = 35\). For zero: \(21 + (n-1)(-3) = 0 \Rightarrow n-1 = 7 \Rightarrow n = 8\). Yes, the 8th term is 0.
Q4. The 4th term of an AP is 0. Show that the 25th term is three times the 11th term.
a + 3d = 0 ⇒ a = −3d. \(a_{11} = a + 10d = -3d + 10d = 7d\). \(a_{25} = a + 24d = -3d + 24d = 21d = 3(7d) = 3 \cdot a_{11}\). ✓
Q5. Find the AP whose 5th term is 19 and whose 9th term exceeds the 4th term by 20.
a + 4d = 19 … (i); (a + 8d) − (a + 3d) = 20 ⇒ 5d = 20 ⇒ d = 4. From (i): a = 19 − 16 = 3. AP: 3, 7, 11, 15, 19, ….
Q6. How many three-digit natural numbers are divisible by 7?
First three-digit multiple of 7 is 105 (= 7 × 15); last is 994 (= 7 × 142). AP: 105, 112, …, 994. n = (994 − 105)/7 + 1 = 889/7 + 1 = 127 + 1 = 128.
Q7. The taxi fare in a city is ₹40 for the first km and ₹15 for each subsequent km. Express the fare for n km as a formula. What is the fare for 12 km?
Fare for first km = 40; the AP of fares per cumulative km is 40, 55, 70, … with d = 15. Total fare for n km = 40 + (n−1)(15) = 25 + 15n. For n = 12: 25 + 180 = ₹205.
Exercise 8.2 — Sum of n Terms
Q8. Find the sum of the first 25 terms of the AP 7, 12, 17, 22, …
a = 7, d = 5, n = 25. \(S_{25} = \tfrac{25}{2}\,[14 + 24(5)] = \tfrac{25}{2}(134) = 25 \times 67 = 1675\).
Q9. Find the sum: 5 + 11 + 17 + … + 95.
a = 5, d = 6, l = 95. n: \(95 = 5 + (n-1)(6) \Rightarrow n − 1 = 15 \Rightarrow n = 16\). \(S = \tfrac{16}{2}(5 + 95) = 8 \times 100 = 800\).
Q10. Find the sum of the first 1000 positive integers.
\(S = \dfrac{1000 \times 1001}{2} = 500{,}500\).
Q11. Find the sum of all multiples of 4 lying between 100 and 500.
Multiples of 4 strictly between 100 and 500: first 104, last 496. n = (496 − 104)/4 + 1 = 99. \(S = \tfrac{99}{2}(104 + 496) = \tfrac{99}{2}(600) = 99 \times 300 = 29{,}700\).
Q12. The sum of the first n terms of an AP is given by \(S_n = 3n^2 + 5n\). Find the AP and its 10th term.
\(a_n = S_n - S_{n-1} = (3n^2+5n) - (3(n-1)^2+5(n-1)) = 6n + 2\). So a₁ = 8, a₂ = 14, a₃ = 20, … d = 6. \(a_{10} = 6(10) + 2 = 62\).
Q13. How many terms of the AP 24, 21, 18, … must be taken so that their sum is 78?
\(\tfrac{n}{2}\,[48 + (n-1)(-3)] = 78 \Rightarrow n[51 - 3n] = 156 \Rightarrow 3n^2 - 51n + 156 = 0 \Rightarrow n^2 - 17n + 52 = 0\). Solve: \(n = \tfrac{17 \pm \sqrt{289 - 208}}{2} = \tfrac{17 \pm 9}{2}\). So n = 13 or n = 4. Both values work: from term 5 onward terms are negative, cancelling out gains.
Exercise 8.3 — Word Problems and Higher-Order
Q14. A man saves ₹100 in the first month and increases his savings by ₹50 each month. (i) How much will he save in the 18th month? (ii) What will be his total savings in 2 years?
(i) a = 100, d = 50, n = 18. \(a_{18} = 100 + 17(50) = 950\). (ii) n = 24. \(S_{24} = \tfrac{24}{2}\,[200 + 23(50)] = 12(1350) = ₹16{,}200\).
Q15. 200 logs are stacked in a heap with 20 logs in the bottom row, 19 in the next, and so on. In how many rows are 200 logs placed and how many logs are in the top row?
a = 20, d = −1, S = 200. \(\tfrac{n}{2}\,[40 - (n-1)] = 200 \Rightarrow n^2 - 41n + 400 = 0\). \(n = \tfrac{41 \pm 9}{2}\) → 25 or 16. n = 25 gives top row of −4 (impossible). So n = 16, top row has \(20 - 15 = 5\) logs.
Q16. In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third and so on. There are 5 plants in the last row. How many rows and how many total plants?
a = 23, d = −2, l = 5. \(5 = 23 + (n-1)(-2) \Rightarrow n - 1 = 9 \Rightarrow n = 10\). \(S = \tfrac{10}{2}(23 + 5) = 5(28) = 140\) plants.
Q17. A spiral is made up of successive semicircles with centers alternating between points A and B, starting with centre A, of radii 0.5, 1.0, 1.5, 2.0, … cm. What is the total length of the spiral made up of 13 consecutive semicircles? (Use π = 22/7.)
Length of n-th semicircle = π × radius_n = π × (n × 0.5). Total length = π × 0.5 × (1 + 2 + … + 13) = π × 0.5 × \(\tfrac{13 \times 14}{2}\) = π × 0.5 × 91 = (22/7) × 45.5 = 143 cm.
Q18. The first and last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
\(350 = 17 + (n-1)(9) \Rightarrow n - 1 = 37 \Rightarrow n = 38\). \(S = \tfrac{38}{2}(17 + 350) = 19 \times 367 = 6973\).
Q19. Find the sum of all even numbers between 1 and 1001.
AP 2, 4, …, 1000. n = 500. \(S = \tfrac{500}{2}(2 + 1000) = 250 \times 1002 = 250{,}500\).
Q20. The interior angles of a polygon are in AP. The smallest angle is 120° and the common difference is 5°. Find the number of sides of the polygon.
Sum of interior angles of n-gon = (n − 2) × 180°. Sum of AP: \(\tfrac{n}{2}\,[240 + (n-1)(5)]\). Equating: \(\tfrac{n}{2}(235 + 5n) = 180(n-2) \Rightarrow 5n^2 + 235n = 360n − 720 \Rightarrow 5n^2 - 125n + 720 = 0 \Rightarrow n^2 - 25n + 144 = 0 \Rightarrow (n-9)(n-16) = 0\). n = 9 or 16. For n = 16, largest angle = 120 + 15(5) = 195° > 180° (not a convex polygon angle), so accept n = 9.
Activity 8.3 — A Class Saving Project
L5 EvaluateMaterials: Notebook, calculator, classmates.
Goal: Use AP knowledge to plan and verify a real saving target.
- Each student picks an amount A to save in week 1 and an increment d for each subsequent week.
- Predict your total saving after 26 weeks (half a year) using \(S_{26} = \tfrac{26}{2}[2A + 25d]\).
- Make a table of your weekly target and weekly cumulative total. Compare with the formula.
- Discuss: which combinations (A, d) lead to large totals quickly? Plot your S₂₆ vs A for fixed d.
If A = 50, d = 10: \(S_{26} = 13 \times (100 + 250) = 13 \times 350 = ₹4{,}550\). If A = 100, d = 5: \(S_{26} = 13 \times (200 + 125) = 13 \times 325 = ₹4{,}225\). The combination with bigger d grows faster in the long run.
Competency-Based Questions
Scenario: A new metro line in a city is being constructed in monthly phases. In month 1, 600 m of track is laid; in each subsequent month, 50 m more is laid than the month before. The full line is 24 months long.
Q1. How much track will be laid in month 12?
L3 ApplyAnswer: (b) 1150 m. a = 600, d = 50, n = 12. \(a_{12} = 600 + 11(50) = 1150\) m.
Q2. The total length of the metro line is the sum of all 24 months. Compute the total length and decide whether a target of 25 km can be met.
L4 AnalyseAnswer: \(S_{24} = \tfrac{24}{2}[1200 + 23(50)] = 12(1200 + 1150) = 12 \times 2350 = 28{,}200\) m = 28.2 km. Since 28.2 km > 25 km, the 25 km target will be met within 24 months.
Q3. The contractor argues, "If we just lay 1175 m every month for 24 months, we get the same total." Evaluate this claim.
L5 EvaluateAnswer: Correct in total. The mean of the AP is \((600 + 1750)/2 = 1175\). 1175 × 24 = 28,200 m = same total. However, the work distribution is different: with constant 1175 m the team works at full pace from day one, while the AP plan has a gentler ramp-up. For practical labour, equipment and budgeting purposes, the AP plan is usually more realistic.
Q4. Design an alternative AP plan starting at the same 600 m in month 1 that will complete exactly 30 km of metro line in 24 months. What common difference is needed?
L6 CreateSolution: a = 600, n = 24, S = 30,000 m. \(30000 = \tfrac{24}{2}\,[1200 + 23d] = 12(1200 + 23d) \Rightarrow 2500 = 1200 + 23d \Rightarrow d = 1300/23 \approx 56.5\) m/month. Round to d = 57 m per month → \(S_{24} = 12(1200 + 23 \times 57) = 12(1200 + 1311) = 30{,}132\) m, slightly above target.
Assertion–Reason Questions
Assertion (A): The AP 2, 5, 8, 11, … has 30th term equal to 89.
Reason (R): The n-th term of an AP is a + (n-1)d.
Reason (R): The n-th term of an AP is a + (n-1)d.
Answer: (a) — \(a_{30} = 2 + 29(3) = 89\). R correctly states the formula used.
Assertion (A): The sum of the first n odd natural numbers is n².
Reason (R): Odd numbers form an AP with a = 1, d = 2; substituting into S = n/2[2a + (n-1)d] gives n².
Reason (R): Odd numbers form an AP with a = 1, d = 2; substituting into S = n/2[2a + (n-1)d] gives n².
Answer: (a) — \(S = \tfrac{n}{2}[2 + (n-1)(2)] = \tfrac{n}{2}(2n) = n^2\). R explains A.
Assertion (A): If a = 5, d = 0, then S_n = 5n for every n.
Reason (R): When d = 0, the AP is constant, so every term equals a and the sum is n × a.
Reason (R): When d = 0, the AP is constant, so every term equals a and the sum is n × a.
Answer: (a) — Both true and R is the precise reason.
Glossary
Key Terms
Sequence: ordered list of numbers.
Term (a_n): the n-th number in a sequence.
Finite / infinite sequence: with or without a last term.
AP (Arithmetic Progression): sequence with constant difference.
Common difference (d): the fixed amount added each step.
n-th term: \(a + (n-1)d\).
Sum of n terms (S_n): \(\tfrac{n}{2}[2a + (n-1)d]\) or \(\tfrac{n}{2}(a + l)\).
Frequently Asked Questions
How do you check if a sequence is an arithmetic progression?
Compute the difference between consecutive terms: a_2 - a_1, a_3 - a_2, a_4 - a_3, .... If all the differences are equal, the sequence is an AP and that common value is d. If any two differences differ, the sequence is not an AP.
If the 5th term of an AP is 17 and the 12th term is 38, find the AP.
From a_5 = a + 4d = 17 and a_12 = a + 11d = 38, subtracting gives 7d = 21, so d = 3. Then a = 17 - 4(3) = 5. The AP is 5, 8, 11, 14, 17, ....
How do you find the sum of the first 30 odd natural numbers?
The first 30 odd numbers form an AP 1, 3, 5, ..., 59 with a = 1, d = 2 and n = 30. S_30 = (30/2)(2(1) + 29(2)) = 15 x (2 + 58) = 15 x 60 = 900. (In general, the sum of the first n odd numbers equals n^2.)
If S_n = 252, a = 3 and d = 2, find n.
Use S_n = (n/2)(2a + (n - 1)d) = (n/2)(6 + 2(n - 1)) = (n/2)(2n + 4) = n^2 + 2n. Set n^2 + 2n = 252, so n^2 + 2n - 252 = 0; solving gives n = 14 (rejecting the negative root). So 14 terms.
A man saves Rs 100 in the first month and Rs 50 more each subsequent month. How much does he save in 12 months?
Monthly savings form an AP with a = 100, d = 50, n = 12. S_12 = (12/2)(2(100) + 11(50)) = 6 x (200 + 550) = 6 x 750 = Rs 4500.
How many three-digit numbers are divisible by 7?
Three-digit multiples of 7 form an AP: 105, 112, 119, ..., 994 with a = 105, d = 7. Use a_n = a + (n - 1)d: 994 = 105 + (n - 1) x 7, so (n - 1) = 127, n = 128. There are 128 such numbers.
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