CBSE Class 9 Maths Ganita Manjari Part 1 Ch 08 Predicting What Comes Next Exploring Sequences and Progressions Long Answer and Case Study Questions with Answers

5 Mark Questions

Question 1. A stadium has 25 seats in the first row, and each next row has 6 more seats than the previous row. There are 18 rows. Find (i) the seats in the last row, (ii) the total seats, and (iii) whether the stadium can seat 1,000 people if every seat is occupied.
Answer:
a = 25, d = 6, n = 18
(i) \( a_{18} = 25 + 17(6) = 127 \) seats
(ii) \( S_{18} = \frac{18}{2}(25 + 127) = 9 \times 152 = 1368 \) seats
(iii) 1368 > 1000, so yes, 1,000 people can be seated, with 368 seats to spare.
In simple words: The last row has 127 seats and the stadium holds 1,368 people in all, more than 1,000.

Exam Tip: In (iii), give the comparison and the spare seats, not just 'yes'.

Question 2. A charity collects ₹500 in week 1 and increases the collection by ₹125 each week for 16 weeks. Find the collection in week 16 and the total collection. If the target is ₹25,000, determine whether the target is met and by how much.
Answer:
a = 500, d = 125, n = 16
\( a_{16} = 500 + 15(125) = 500 + 1875 = ₹2,375 \)
\( S_{16} = \frac{16}{2}(500 + 2375) = 8 \times 2875 = ₹23,000 \)
The target of ₹25,000 is not met; it falls short by ₹2,000.
In simple words: Week 16 brings ₹2,375, but the 16-week total of ₹23,000 is ₹2,000 below the target.

Exam Tip: 'By how much' means find the difference: 25,000 − 23,000.

Question 3. In an AP, the sum of the first 10 terms is 230 and the 10th term is 41. Find the first term and common difference. Then find the 20th term.
Answer:
\( S_{10} = \frac{10}{2}(a + a_{10}) \Rightarrow 230 = 5(a + 41) \Rightarrow a + 41 = 46 \Rightarrow a = 5 \)
\( a_{10} = a + 9d \Rightarrow 41 = 5 + 9d \Rightarrow d = 4 \)
\( a_{20} = 5 + 19(4) = 81 \)
a = 5, d = 4, 20th term = 81.
In simple words: The sum formula with the last term gives the first term straight away, and then d follows.

Exam Tip: \( S_n = \frac{n}{2}(a + l) \) is ideal when the last term is given.

Question 4. A game awards points in levels according to 7, 12, 17, … . A player completes 25 levels. Find the total points earned. If the player needs at least 2,000 points, how many additional complete levels are needed?
Answer:
a = 7, d = 5, n = 25
\( S_{25} = \frac{25}{2}[2(7) + 24(5)] = \frac{25}{2} \times 134 = 1675 \) points
Points still needed = 2000 − 1675 = 325
Next levels: \( a_{26} = 7 + 25(5) = 132 \), \( a_{27} = 137 \), \( a_{28} = 142 \)
After 2 more levels: 132 + 137 = 269 < 325. After 3 more levels: 269 + 142 = 411 ≥ 325.
Total after 25 levels: 1,675 points. The player needs 3 more complete levels (reaching 2,086 points).
In simple words: 25 levels give 1,675 points, and the next three levels (132 + 137 + 142) push the total past 2,000.

Exam Tip: Add the new levels one at a time and stop as soon as the running total reaches the target.

Case Study Based Questions

Question 5. A school creates a stepped flower bed. The first row contains 9 plants, the second 13, the third 17, and so on.
(a) Identify the first term and common difference.
(b) Find the number of plants in the 12th row.
(c) Find the total number of plants in the first 12 rows.
Answer:
(a) First term a = 9; common difference d = 13 − 9 = 4
(b) \( a_{12} = 9 + 11(4) = 53 \) plants
(c) \( S_{12} = \frac{12}{2}(9 + 53) = 6 \times 62 = 372 \) plants
In simple words: Each row has 4 more plants; the 12th row has 53, and the 12 rows hold 372 plants in all.

Exam Tip: Use the answer to (b) as the last term in (c).

Question 6. A mobile learning app awards 20 points on the first day of a challenge and increases the daily award by 10 points. A learner completes 15 days.
(a) Find the points on day 15.
(b) Find the total points earned in 15 days.
(c) The learner needs 1,500 points. Does the learner reach the target? If not, how many more points are needed?
Answer:
(a) \( a_{15} = 20 + 14(10) = 160 \) points
(b) \( S_{15} = \frac{15}{2}(20 + 160) = \frac{15}{2} \times 180 = 1350 \) points
(c) 1350 < 1500, so the target is not reached. 150 more points are needed.
In simple words: Day 15 gives 160 points, the total is 1,350, and the learner is 150 points short.

Exam Tip: Always compare the total with the target and state the shortfall.

Question 7. A theatre has 30 seats in the first row, 34 in the second, 38 in the third, and the pattern continues for 20 rows.
(a) Find the number of seats in the 20th row.
(b) Find the total number of seats.
(c) If 750 people arrive, can everyone be seated? Justify using the total.
Answer:
(a) \( a_{20} = 30 + 19(4) = 106 \) seats
(b) \( S_{20} = \frac{20}{2}(30 + 106) = 10 \times 136 = 1360 \) seats
(c) 1360 > 750, so yes, everyone can be seated, with 610 seats left over.
In simple words: The theatre holds 1,360 people, far more than the 750 who arrive.

Exam Tip: Justify with the numbers: 1360 − 750 = 610 spare seats.