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

2 Mark Questions

Question 1. A pattern begins 14, 19, 24, 29, … . Find the 15th term and explain the step used to obtain it.
Answer:
a = 14 and d = 19 − 14 = 5.
\( a_{15} = a + 14d = 14 + 14(5) = 14 + 70 = 84 \)
The 15th term is 84. We add the common difference 14 times (not 15), because the first term is already there.
In simple words: Start at 14 and add 5 fourteen times to reach the 15th term.

Exam Tip: Use \( a_n = a + (n - 1)d \); the (n − 1) is where most errors happen.

Question 2. The 7th term of an AP is 31 and its common difference is 4. Find its first term.
Answer:
\( a_7 = a + 6d \Rightarrow 31 = a + 6(4) = a + 24 \)
\( a = 7 \)
In simple words: Go back 6 steps of 4 from 31 to reach the first term.

Exam Tip: Check: 7, 11, 15, 19, 23, 27, 31 ✓.

Question 3. A delivery service charges ₹60 on day 1 and increases the charge by ₹15 each day. Find the charge on day 12.
Answer:
a = 60, d = 15 and n = 12.
\( a_{12} = 60 + 11(15) = 60 + 165 = 225 \)
The charge on day 12 is ₹225.
In simple words: From day 1 to day 12 there are 11 increases of ₹15.

Exam Tip: Write the unit (₹) in the final answer.

Question 4. For the AP 7, 12, 17, …, which term is 92? Show your reasoning.
Answer:
a = 7 and d = 5. Solve \( 7 + (n - 1)5 = 92 \).
\( (n - 1)5 = 85 \Rightarrow n - 1 = 17 \Rightarrow n = 18 \)
92 is the 18th term.
In simple words: 92 is 85 more than 7, which is 17 steps of 5, so it is the 18th term.

Exam Tip: If n comes out as a fraction or negative, the number is not a term of the AP.

Question 5. A sequence has nth term \( 5n - 1 \). Find the 20th term and the difference between two consecutive terms.
Answer:
\( a_{20} = 5(20) - 1 = 99 \)
\( a_{n+1} - a_n = [5(n + 1) - 1] - [5n - 1] = 5 \)
20th term = 99; consecutive terms differ by 5.
In simple words: Put n = 20 to get 99, and the coefficient of n (5) is the common difference.

Exam Tip: In any rule of the form \( an + b \), the common difference is a.

3 Mark Questions

Question 6. A school auditorium has 18 seats in the first row, 22 in the second, 26 in the third, and so on. Find the number of seats in the 15th row and the total seats in the first 15 rows.
Answer:
a = 18, d = 4, n = 15
\( a_{15} = 18 + 14(4) = 74 \) seats
\( S_{15} = \frac{15}{2}(18 + 74) = \frac{15}{2} \times 92 = 690 \) seats
15th row: 74 seats; first 15 rows: 690 seats.
In simple words: The 15th row has 74 seats, and the average row (18 + 74) ÷ 2 = 46 seats times 15 rows gives 690.

Exam Tip: Once you know the last term, \( S_n = \frac{n}{2}(a + l) \) is the quickest sum formula.

Question 7. A student saves ₹200 in the first month and increases the saving by ₹75 every month. How much is saved in the 12th month? Also find the total saving in 12 months.
Answer:
a = 200, d = 75, n = 12
\( a_{12} = 200 + 11(75) = 200 + 825 = 1025 \)
\( S_{12} = \frac{12}{2}(200 + 1025) = 6 \times 1225 = 7350 \)
12th month: ₹1,025; total in 12 months: ₹7,350.
In simple words: The saving grows to ₹1,025 by month 12, and the 12 months add up to ₹7,350.

Exam Tip: Keep the two answers separate: one term versus the sum of all terms.

Question 8. The 4th term of an AP is 17 and the 10th term is 41. Find the first term, common difference, and 20th term.
Answer:
\( a + 3d = 17 \) … (1) and \( a + 9d = 41 \) … (2)
(2) − (1): \( 6d = 24 \Rightarrow d = 4 \)
From (1): \( a = 17 - 12 = 5 \)
\( a_{20} = 5 + 19(4) = 81 \)
a = 5, d = 4, 20th term = 81.
In simple words: The jump from the 4th to the 10th term (24) spread over 6 steps gives d = 4.

Exam Tip: Subtracting the two equations removes a and gives d directly.

Question 9. A sequence has first term 35 and common difference −4. Determine whether 3 is a term of the sequence. Give a reason.
Answer:
\( a_n = 35 + (n - 1)(-4) \). Set \( a_n = 3 \):
\( 35 - 4(n - 1) = 3 \Rightarrow 4(n - 1) = 32 \Rightarrow n - 1 = 8 \Rightarrow n = 9 \)
Since n = 9 is a natural number, 3 is a term: it is the 9th term.
In simple words: Counting down in 4s from 35 reaches 3 after 8 steps, so 3 is the 9th term.

Exam Tip: A number is a term only if the value of n comes out as a positive whole number.

Question 10. The numbers of books read by students form the AP 3, 7, 11, 15, … . How many books would the 20th student have read according to the pattern? What is the total for the first 20 students?
Answer:
a = 3, d = 4, n = 20
\( a_{20} = 3 + 19(4) = 79 \) books
\( S_{20} = \frac{20}{2}(3 + 79) = 10 \times 82 = 820 \) books
20th student: 79 books; first 20 students: 820 books.
In simple words: The 20th student reads 79 books, and all 20 together read 820.

Exam Tip: Find the last term first, then use \( S_n = \frac{n}{2}(a + l) \).

Question 11. A designer makes triangular arrangements using 1, 3, 6, 10, 15, … tiles. The differences are 2, 3, 4, 5, … . Is this sequence an AP? Explain, and predict the next two terms.
Answer:
No, it is not an AP, because the differences (2, 3, 4, 5, …) are not constant.
The differences themselves increase by 1 each time, so the next differences are 6 and 7.
Next terms: 15 + 6 = 21 and 21 + 7 = 28.
(These are the triangular numbers \( \frac{n(n + 1)}{2} \).)
In simple words: The gaps keep growing, so it is not an AP; the next gaps of 6 and 7 give 21 and 28.

Exam Tip: An AP needs one fixed difference; growing differences mean some other kind of pattern.