Arithmetic Progressions — Question Bank

Class X · Mathematics Ch.5 · CBSE Pattern · 30 Questions

10
MCQ (1 mark)
3
Assertion-Reason
12
Short Answer (2-3 m)
5
Long Answer (5 m)
47
Total Marks
Section A — Multiple Choice Questions (1 Mark Each)
MCQ 1
Which of the following lists of numbers forms an AP?
MCQ 2
The 10th term of the AP 2, 7, 12, … is:
MCQ 3
The common difference of the AP 10, 7, 4, 1, … is:
MCQ 4
The sum of the first n natural numbers is given by:
MCQ 5
Which term of the AP 3, 8, 13, 18, … is 78?
MCQ 6
The common difference of a constant AP such as 5, 5, 5, … is:
MCQ 7
The sum of the first 20 terms of the AP 2, 5, 8, … is:
MCQ 8
If aₙ = Sₙ − Sₙ₋₁ and Sₙ = 3n² + 5n, then a₁ equals:
MCQ 9
Three numbers in AP are best represented as:
MCQ 10
How many two-digit numbers are divisible by 3?
Section B — Assertion & Reason (1 Mark Each)
A-R 1
Choose the correct option about the statements below.
ASSERTION (A)The sequence 5, 8, 11, 14, … is an AP.
REASON (R)A sequence is an AP if the difference between consecutive terms is constant.
A-R 2
Choose the correct option about the statements below.
ASSERTION (A)The number 100 is a term of the AP 5, 11, 17, …
REASON (R)A value is a term of an AP only if aₙ = a+(n−1)d gives a positive integer n.
A-R 3
Choose the correct option about the statements below.
ASSERTION (A)If a constant k is added to each term of an AP, the result is still an AP.
REASON (R)Adding a constant to every term changes the common difference.
Section C — Short Answer Questions (2–3 Marks Each)
SA 1
Check whether −150 is a term of the AP 11, 8, 5, 2, …
[2 marks]
a = 11, d = 8−11 = −3.
aₙ = 11 + (n−1)(−3) = 14 − 3n.
Set 14 − 3n = −150 → 3n = 164 → n = 54.67…

Since n is not a positive integer, −150 is not a term of this AP.
SA 2
The 5th term of an AP is 19 and the 8th term is 31. Find the AP.
[3 marks]
a + 4d = 19 … (i)
a + 7d = 31 … (ii)

Subtract (i) from (ii): 3d = 12 → d = 4.
From (i): a = 19 − 16 = 3.

The AP is 3, 7, 11, 15, …
SA 3
Find the 20th term from the end of the AP 3, 8, 13, …, 253.
[2 marks]
a = 3, d = 5, l = 253.
nth term from end = l − (n−1)d.
20th from end = 253 − (20−1)×5 = 253 − 95 = 158.
SA 4
Find the sum: 7 + 10½ + 14 + … + 84.
[3 marks]
a = 7, d = 3.5, l = 84.
84 = 7 + (n−1)(3.5) → 77 = 3.5(n−1) → n−1 = 22 → n = 23.

Sₙ = n/2 (a + l) = 23/2 (7 + 84) = 23/2 × 91 = 1046.5.
SA 5
For what value of k are the numbers k+2, 4k−6, 3k−2 in AP?
[2 marks]
In an AP, middle term = average of neighbours, so:
2(4k−6) = (k+2) + (3k−2)
8k − 12 = 4k
4k = 12 → k = 3.

Check: terms become 5, 6, 7 — an AP with d = 1 ✓
SA 6
How many terms of the AP 18, 16, 14, … are needed to give a sum of 78?
[3 marks]
a = 18, d = −2, Sₙ = 78.
78 = n/2[36 + (n−1)(−2)] = n/2[38 − 2n] = n(19 − n)
n² − 19n + 78 = 0 → (n−6)(n−13) = 0
n = 6 or n = 13.

Both are valid (terms 7–13 sum to 0), so the sum is 78 at either 6 or 13 terms.
SA 7
Find the middle term of the AP 7, 13, 19, …, 247.
[2 marks]
a = 7, d = 6, l = 247.
247 = 7 + (n−1)6 → n−1 = 40 → n = 41 terms (odd).
Middle term = the (41+1)/2 = 21st term.
a₂₁ = 7 + 20×6 = 127.
SA 8
A man saves ₹200 in the first month and increases his savings by ₹50 each month. How much does he save in the 12th month?
[2 marks]
a = 200, d = 50.
a₁₂ = 200 + (12−1)×50 = 200 + 550 = ₹750.
SA 9
The first term of an AP is 5, the last term is 45 and the sum of all terms is 400. Find the number of terms and the common difference.
[3 marks]
Sₙ = n/2 (a + l) → 400 = n/2 (5 + 45) = 25n → n = 16.

l = a + (n−1)d → 45 = 5 + 15d → 15d = 40 → d = 8/3.
SA 10
If Sₙ = 4n − n², find the first term, the second term and the nth term.
[3 marks]
S₁ = 4 − 1 = 3 → a₁ = 3.
S₂ = 8 − 4 = 4 → a₂ = S₂ − S₁ = 4 − 3 = 1 → d = −2.

aₙ = Sₙ − Sₙ₋₁ = (4n − n²) − [4(n−1) − (n−1)²]
= 4n − n² − (4n − 4 − n² + 2n − 1) = 5 − 2n.
SA 11
Determine the AP whose 3rd term is 5 and 7th term is 9.
[2 marks]
a + 2d = 5, a + 6d = 9. Subtract: 4d = 4 → d = 1.
a = 5 − 2 = 3.
AP: 3, 4, 5, 6, …
SA 12
In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. If there are 5 plants in the last row, how many rows are there?
[3 marks]
a = 23, d = −2, last term l = 5.
5 = 23 + (n−1)(−2)
−18 = −2(n−1) → n−1 = 9 → n = 10 rows.
Section D — Long Answer Questions (5 Marks Each)
LA 1
The sum of three numbers in AP is 27 and the sum of their squares is 293. Find the numbers.
[5 marks]
Let the numbers be a−d, a, a+d.
Sum: (a−d)+a+(a+d) = 3a = 27 → a = 9.

Sum of squares: (a−d)² + a² + (a+d)² = 293
3a² + 2d² = 293 → 3(81) + 2d² = 293
2d² = 293 − 243 = 50 → d² = 25 → d = ±5.

Numbers are 4, 9, 14 (or 14, 9, 4).
LA 2
Find the sum of all three-digit numbers which are divisible by 7.
[5 marks]
Smallest 3-digit multiple of 7 = 105; largest = 994.
AP: 105, 112, …, 994, with d = 7.

994 = 105 + (n−1)7 → 889 = 7(n−1) → n−1 = 127 → n = 128.

Sₙ = n/2 (a + l) = 128/2 (105 + 994) = 64 × 1099 = 70,336.
LA 3
200 logs are stacked so that there are 20 logs in the bottom row, 19 in the next, 18 in the row above and so on. In how many rows are the 200 logs placed, and how many logs are in the top row?
[5 marks]
a = 20, d = −1, Sₙ = 200.
200 = n/2[40 + (n−1)(−1)] = n/2[41 − n]
400 = 41n − n² → n² − 41n + 400 = 0
(n − 16)(n − 25) = 0 → n = 16 or 25.

For n = 25: a₂₅ = 20 + 24(−1) = −4 (impossible — reject).
So n = 16 rows. Top row: a₁₆ = 20 + 15(−1) = 5 logs.
LA 4
A sum of ₹700 is to be used to give seven cash prizes to students. If each prize is ₹20 less than the preceding prize, find the value of each of the prizes.
[5 marks]
Let the largest prize = a, d = −20, n = 7, S₇ = 700.
700 = 7/2 [2a + 6(−20)] = 7/2 (2a − 120)
200 = 2a − 120 → 2a = 320 → a = 160.

Prizes: ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40.
LA 5
If the sum of the first 7 terms of an AP is 49 and that of the first 17 terms is 289, find the sum of its first n terms.
[5 marks]
S₇ = 7/2[2a + 6d] = 7(a + 3d) = 49 → a + 3d = 7 … (i)
S₁₇ = 17/2[2a + 16d] = 17(a + 8d) = 289 → a + 8d = 17 … (ii)

Subtract: 5d = 10 → d = 2; from (i) a = 7 − 6 = 1.

Sₙ = n/2[2(1) + (n−1)2] = n/2[2n] = n².