Sequences and Series — Chapter Test | Class 11 IMO

Q1

The sum of n terms of the series 5 + 55 + 555 + ... is:

S = 5(1 + 11 + 111 + ...) = (5/9)(9 + 99 + 999 + ...) = (5/9)[(10−1) + (10²−1) + ...] = (5/9)[10(10ⁿ−1)/9 − n].

Q2

The sum of the infinite GP 3 + 1 + 1/3 + ... is:

S∞ = 3/(1−1/3)=4.5.

Q3

If x, y, z are positive real numbers, the minimum value of (x+y)(y+z)(z+x) is:

Using AM ≥ GM for each pair: x+y ≥ 2√(xy), y+z ≥ 2√(yz), z+x ≥ 2√(zx). Multiplying gives (x+y)(y+z)(z+x) ≥ 8√(x²y²z²) = 8xyz.

Q4

The common ratio of the GP 2, -6, 18, -54, ... is:

r = -6/2 = -3.

Q5

The 13th term of the AP 2, 7, 12, ... is:

T₁₃ = 2 + 12×5 = 62.

Q6

A rubber ball is dropped from a height of 10 metres. Each time it bounces, it rises to 3/4 of its previous height. The total distance covered by the ball before coming to rest is:

Distance = 10 + 2 × (10 × 3/4) + 2 × (10 × (3/4)²) + ... = 10 + 15 / (1 − 3/4) = 10 + 15 / (1/4) = 10 + 60 = 70 m.

Q7

The sum of the first 15 odd natural numbers is:

Sum of first n odd numbers = n² = 225.

Q8

If x + 2, 3x, and 4x + 1 are in AP, then the value of x is:

For terms to be in AP, 2b = a + c. So, 2(3x) = (x + 2) + (4x + 1) → 6x = 5x + 3 → x = 3.

Q9

Evaluate the telescoping series 1/(1×2) + 1/(2×3) + 1/(3×4) + ... + 1/(9×10).

Each term 1/[n(n+1)] can be split into 1/n − 1/(n+1). The series becomes (1 − 1/2) + (1/2 − 1/3) + ... + (1/9 − 1/10) = 1 − 1/10 = 9/10.

Q10

A viral message is sent to 4 people. Each of those sends it to 4 new people, and so on. How many people receive the message by the 5th level?

This is a GP: 4 + 16 + 64 + 256 + 1024. Sum S5 = 4(4⁵ − 1)/(4 − 1) = 4(1023)/3 = 4(341) = 1364.

Q11

A new motorcycle costs ₹ 1,00,000 and depreciates by 10% each year. What will be its value at the end of 3 years?

Value after 1 year = 1,00,000 × 0.9. After 3 years = 1,00,000 × (0.9)³ = 1,00,000 × 0.729 = 72,900.

Q12

What is the nth term of the series 1, 4, 9, 16, 25, ...?

The terms are squares of natural numbers: 1², 2², 3², 4², 5²... Therefore, the nth term is n².

Q13

The arithmetic mean of 7 and 19 is:

(7+19)/2 = 13.

Q14

Find the 10th term of the series 3, 7, 13, 21, 31, ... using the method of differences.

Differences: 4, 6, 8, 10 (an AP). tⁿ = a + sum of (n-1) differences = 3 + (n-1)/2[2(4) + (n-2)2] = n² + n + 1. For n=10, 100 + 10 + 1 = 111.

Q15

If a, b, c are in AP, then a³, b³, c³ are in:

Consider the AP 1, 2, 3. Their cubes are 1, 8, 27. 8−1=7, 27−8=19. The differences are not constant, and there is no constant ratio. They form none of these standard progressions.