RowQ
The Vault
RowQ
The Vault
CBSE Class 10 Maths · 11 questions · 26 marks
Some lists of numbers grow by the same fixed step every time — salaries rising by a fixed increment, seats increasing row by row, a stack of pipes narrowing upwards. Such a list is an arithmetic progression, and two formulas describe it completely: one for the nth term and one for the sum of the first n terms. Once you can spot a and d, almost every question in this chapter is a substitution away from being finished.
The 11th term of the AP 7, 11, 15, 19, … is:
Answer
The 11th term is 47. Here a = 7 and d = 11 - 7 = 4. a₁₁ = a + (11 - 1)d = 7 + 10(4) = 7 + 40 = 47. A common slip is to use 11d instead of 10d, which would wrongly give 51.
Which term of the AP 21, 18, 15, 12, … is -30?
Answer
It is the 18th term. Here a = 21 and d = 18 - 21 = -3. Set aₙ = -30: 21 + (n - 1)(-3) = -30. -3(n - 1) = -51, so n - 1 = 17 and n = 18. Check: a₁₈ = 21 + 17(-3) = 21 - 51 = -30. ✔
The sum of the first 15 terms of an AP whose first term is 5 and common difference is 3 is:
Answer
The sum is 390. Sₙ = (n/2)[2a + (n - 1)d] with n = 15, a = 5 and d = 3. S₁₅ = (15/2)[2(5) + 14(3)] = (15/2)[10 + 42] = (15/2)(52) = 15 × 26 = 390. As a check, the 15th term is 5 + 14(3) = 47, and (15/2)(5 + 47) = (15/2)(52) = 390.
In an AP, the 8th term exceeds the 3rd term by 20. The common difference is:
Answer
The common difference is 4. a₈ = a + 7d and a₃ = a + 2d. So a₈ - a₃ = (a + 7d) - (a + 2d) = 5d. Given 5d = 20, we get d = 4. Notice that the first term a is not needed and in fact cannot be determined from this information alone.
Assertion (A): The sequence 2, 6, 12, 20, 30, … is not an arithmetic progression. Reason (R): A sequence is an AP only if the difference between every pair of consecutive terms is the same.
Answer
Both A and R are true and R is the correct explanation of A. Reason R is the definition of an AP and is true. The successive differences here are 6 - 2 = 4, 12 - 6 = 6, 20 - 12 = 8 and 30 - 20 = 10. These differences are not constant, so by the criterion in R the sequence is not an AP, which is Assertion A. Hence R explains A.
How many terms are there in the AP 9, 14, 19, 24, …, 154?
Answer
Here a = 9, d = 14 - 9 = 5 and the last term aₙ = 154. Using aₙ = a + (n - 1)d: 154 = 9 + (n - 1)(5) 145 = 5(n - 1), so n - 1 = 29 and n = 30. The AP has 30 terms.
The 5th term of an AP is 19 and its 12th term is 47. Find the first term, the common difference and the 20th term.
Answer
Using aₙ = a + (n - 1)d: a + 4d = 19 ...(i) a + 11d = 47 ...(ii) Subtract (i) from (ii): 7d = 28, so d = 4. Substitute in (i): a + 16 = 19, so a = 3. The AP begins 3, 7, 11, 15, 19, … a₂₀ = 3 + 19(4) = 3 + 76 = 79. So the first term is 3, the common difference is 4 and the 20th term is 79.
Find the sum of the first 25 terms of the AP 4, 9, 14, 19, …
Answer
Here a = 4, d = 5 and n = 25. S₂₅ = (25/2)[2(4) + (25 - 1)(5)] = (25/2)[8 + 120] = (25/2)(128) = 25 × 64 = 1600. The sum of the first 25 terms is 1600.
How many terms of the AP 27, 24, 21, 18, … must be added so that their sum is 135? Explain the two answers you obtain.
Answer
Here a = 27 and d = 24 - 27 = -3. Using Sₙ = (n/2)[2a + (n - 1)d]: 135 = (n/2)[54 + (n - 1)(-3)] 135 = (n/2)[54 - 3n + 3] = (n/2)(57 - 3n) 270 = n(57 - 3n) = 57n - 3n² 3n² - 57n + 270 = 0, and dividing by 3, n² - 19n + 90 = 0. Factorise: n² - 10n - 9n + 90 = n(n - 10) - 9(n - 10) = (n - 9)(n - 10) = 0. So n = 9 or n = 10. Both answers are genuine. The 10th term is a₁₀ = 27 + 9(-3) = 0, and adding a term equal to zero does not change the sum. So the sum of the first 9 terms and the sum of the first 10 terms are both 135. Verification: S₉ = (9/2)[54 + 8(-3)] = (9/2)(30) = 135 ✔ and S₁₀ = (10/2)[54 + 9(-3)] = 5(27) = 135 ✔
A teacher saves ₹ 3200 in the first year of her job and increases her yearly saving by ₹ 400 every year. (a) In how many years will her total savings amount to ₹ 50000? (b) How much does she save in that final year? (c) How much has she saved in the 6th year?
Answer
(a) The yearly savings form an AP with a = 3200 and d = 400. Sₙ = (n/2)[2(3200) + (n - 1)(400)] = (n/2)[6400 + 400n - 400] = (n/2)(6000 + 400n) = 3000n + 200n². Set this equal to 50000: 200n² + 3000n - 50000 = 0. Divide by 200: n² + 15n - 250 = 0. D = 15² - 4(1)(-250) = 225 + 1000 = 1225, and √1225 = 35. n = [-15 ± 35]/2, so n = 10 or n = -25. The number of years cannot be negative, so n = 10. Her total savings reach ₹ 50000 in 10 years. (b) Saving in the 10th year = a₁₀ = 3200 + 9(400) = 3200 + 3600 = ₹ 6800. (c) Saving in the 6th year = a₆ = 3200 + 5(400) = 3200 + 2000 = ₹ 5200. Check of (a): S₁₀ = (10/2)(3200 + 6800) = 5 × 10000 = 50000 ✔
Read the following and answer the questions that follow: A school auditorium is being built with 20 rows of seats. The first row has 18 seats and every following row has 3 seats more than the row in front of it. (a) How many seats are there in the 10th row? (b) Which row has 66 seats? (c) Find the total number of seats in the auditorium. (d) How many seats are there in the last five rows taken together?
Answer
The numbers of seats form an AP with a = 18 and d = 3. (a) a₁₀ = 18 + 9(3) = 18 + 27 = 45 seats. (b) Set aₙ = 66: 18 + (n - 1)(3) = 66, so 3(n - 1) = 48, giving n - 1 = 16 and n = 17. The 17th row has 66 seats. (c) S₂₀ = (20/2)[2(18) + 19(3)] = 10[36 + 57] = 10 × 93 = 930 seats in all. (d) The last five rows are rows 16 to 20. S₁₅ = (15/2)[36 + 14(3)] = (15/2)(78) = 15 × 39 = 585. Seats in the last five rows = S₂₀ - S₁₅ = 930 - 585 = 345. Check directly: a₁₆ = 18 + 15(3) = 63 and a₂₀ = 18 + 19(3) = 75, so the sum is (5/2)(63 + 75) = (5/2)(138) = 345. ✔
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