Arithmetic Progressions — Chapter Test | Class 10 IMO

Q1

A taxi charges ₹15 for the first km and ₹8 for each additional km. The fare for 10 km is:

Fare = 15 + 9 × 8 = ₹87.

Q2

The first negative term of the AP 20, 17, 14, … is:

aₙ = 20 − 3(n − 1) < 0 first at n = 8, where a₈ = 20 − 21 = −1.

Q3

The sum of all multiples of 3 from 3 to 30 is:

There are 10 terms; sum = 10/2 (3 + 30) = 5 × 33 = 165.

Q4

The arithmetic mean of 7 and 23 is:

Mean = (7 + 23)/2 = 15.

Q5

The 10th term of the AP 2, 5, 8, … is:

a₁₀ = a + 9d = 2 + 9 × 3 = 29.

Q6

The sequence of even natural numbers forms an AP with common difference:

2, 4, 6, 8, … increases by 2.

Q7

If 7 times the 7th term of an AP equals 11 times its 11th term, then the 18th term is:

7(a + 6d) = 11(a + 10d) gives 4a + 68d = 0, i.e. a + 17d = 0, which is the 18th term, so it is 0.

Q8

Which sequence is NOT an AP?

3, 6, 12 doubles each time, so it has no constant difference.

Q9

If three numbers are in AP, the middle number equals:

For a, b, c in AP, 2b = a + c, so b is the average.

Q10

Reading the terms of an AP from last to first gives:

Reversing an AP gives another AP whose common difference is the negative of the original.

Q11

A worker's starting salary is ₹8000, rising by ₹500 each year. In the 10th year his salary is:

a₁₀ = 8000 + 9 × 500 = ₹12500.

Q12

If the sum of the first n terms of an AP is Sₙ = 2n² + 3n, the common difference is:

aₙ = Sₙ − Sₙ₋₁ = (2n² + 3n) − (2(n−1)² + 3(n−1)) = 4n + 1, so d = 4.

Q13

If the first term of an AP is 7 and its 13th term is 43, the common difference is:

43 = 7 + 12d gives 12d = 36, so d = 3.

Q14

A clock strikes 1 at one o'clock, 2 at two o'clock, and so on up to 12. In twelve hours it strikes a total of:

1 + 2 + … + 12 = 12 × 13 ÷ 2 = 78 times.

Q15

The 15th term of an AP with a = 4 and d = 3 is:

a₁₅ = 4 + 14 × 3 = 46.