RowQ
The Vault
RowQ
The Vault
CBSE Class 9 Maths · 11 questions · 26 marks
Numbers often arrive in an orderly queue: seats increasing row by row in an auditorium, or savings growing by a fixed amount each month. This chapter gives you the language of terms, common differences and common ratios so you can jump straight to the 40th term without listing the first 39. Sums then follow from a neat pairing trick that is worth understanding rather than memorising.
The common difference of the AP 17, 12, 7, 2, ... is:
Answer
The common difference is -5. The common difference d is any term minus the term before it: 12 - 17 = -5, and checking again, 7 - 12 = -5. Because d is negative the sequence is decreasing, which matches the pattern given.
The 15th term of the AP whose first term is 6 and whose common difference is 4 is:
Answer
The 15th term is 62. Using aₙ = a + (n - 1)d with a = 6, d = 4 and n = 15: a₁₅ = 6 + 14 × 4 = 6 + 56 = 62. A common slip is to multiply by 15 instead of (n - 1) = 14, which would wrongly give 66.
For the GP 3, 6, 12, 24, ..., the 7th term is:
Answer
The 7th term is 192. Here a = 3 and r = 6/3 = 2. Using aₙ = a·r^(n-1), a₇ = 3 × 2⁶ = 3 × 64 = 192. Writing the terms out confirms this: 3, 6, 12, 24, 48, 96, 192.
The sum of the first 20 natural numbers is:
Answer
The sum is 210. Using the formula 1 + 2 + ... + n = n(n + 1)/2 with n = 20 gives 20 × 21/2 = 420/2 = 210. The result 420 comes from forgetting to halve, since pairing 1 with 20, 2 with 19 and so on gives 10 pairs each summing to 21.
Assertion (A): The sequence 4, 8, 16, 32, ... is an arithmetic progression. Reason (R): In an arithmetic progression the difference between any two consecutive terms is constant.
Answer
A is false but R is true. Reason R correctly states the defining property of an AP. Checking the sequence, 8 - 4 = 4 but 16 - 8 = 8, so the differences are not constant and the sequence is not an AP. It is in fact a GP with common ratio 2, since each term is twice the one before. Hence A is false while R is true.
Which term of the AP 9, 14, 19, 24, ... is 129?
Answer
Here a = 9 and d = 14 - 9 = 5. Set aₙ = 129: 9 + (n - 1)5 = 129. So (n - 1)5 = 120, giving n - 1 = 24 and n = 25. Hence 129 is the 25th term of the AP.
The 5th term of an AP is 23 and the 12th term is 58. Find the first term and the common difference.
Answer
Using aₙ = a + (n - 1)d: a + 4d = 23 ...(i) a + 11d = 58 ...(ii) Subtract (i) from (ii): 7d = 35, so d = 5. Substitute d = 5 into (i): a + 20 = 23, so a = 3. The first term is 3 and the common difference is 5, giving the AP 3, 8, 13, 18, 23, ...
Three numbers in AP have a sum of 51 and the smallest of them is 13. Find the other two numbers.
Answer
Let the three numbers be a - d, a and a + d. Their sum is 3a = 51, so a = 17. The smallest number is a - d = 13, so 17 - d = 13, giving d = 4. The three numbers are 13, 17 and 21, and the two required numbers are 17 and 21.
A community library adds books to its shelves in an arithmetic pattern: 48 books in the first month, 55 in the second, 62 in the third, and so on. (a) How many books are added in the 18th month? (b) What is the total number of books added in the first 18 months? (c) In which month does the monthly addition first exceed 150 books?
Answer
Here a = 48 and d = 55 - 48 = 7. (a) a₁₈ = a + 17d = 48 + 17 × 7 = 48 + 119 = 167 books. (b) S₁₈ = (n/2)(a + l) = (18/2)(48 + 167) = 9 × 215 = 1935 books in total. (c) We need 48 + (n - 1)7 > 150, so (n - 1)7 > 102 and n - 1 > 14.57..., giving n > 15.57... The smallest whole number satisfying this is n = 16. Checking: a₁₆ = 48 + 15 × 7 = 153, which exceeds 150, while a₁₅ = 146 does not. So the addition first exceeds 150 books in the 16th month.
(a) Find the sum of all three-digit multiples of 7. (b) Find the sum of the first 8 terms of the GP 5, 10, 20, 40, ...
Answer
(a) The smallest three-digit multiple of 7 is 105 (7 × 15) and the largest is 994 (7 × 142). These form an AP with a = 105, d = 7 and l = 994. Find n: 994 = 105 + (n - 1)7, so (n - 1)7 = 889, n - 1 = 127 and n = 128. Sum = (n/2)(a + l) = (128/2)(105 + 994) = 64 × 1099 = 70336. (b) For the GP, a = 5 and r = 10/5 = 2, with n = 8. Sₙ = a(rⁿ - 1)/(r - 1) = 5(2⁸ - 1)/(2 - 1) = 5(256 - 1)/1 = 5 × 255 = 1275. So the required sums are 70336 and 1275.
Read the following and answer the questions that follow: An open-air theatre is built with rows of seats arranged so each row holds 4 more seats than the row in front of it. The front row holds 22 seats, and the theatre has 25 rows in all. (a) Explain why the seat counts form an AP and state a and d. (b) How many seats are in the last (25th) row? (c) What is the total seating capacity of the theatre? (d) A block of the first 10 rows is reserved for guests. How many seats are reserved?
Answer
(a) Each row has exactly 4 more seats than the previous one, so the difference between consecutive terms is constant. That constant difference is the defining property of an AP. Here the first term a = 22 and the common difference d = 4. (b) a₂₅ = a + 24d = 22 + 24 × 4 = 22 + 96 = 118 seats in the last row. (c) S₂₅ = (25/2)(a + l) = (25/2)(22 + 118) = (25/2)(140) = 25 × 70 = 1750 seats in total. (d) S₁₀ = (10/2)[2a + 9d] = 5[44 + 36] = 5 × 80 = 400 seats reserved for guests.
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