IMO Maths Olympiad Class 10 Chapter 05 Arithmetic Progressions MCQs with Answers

Answer: (b) −490

Explanation:
1. T₁ = 7 − 3 = 4, T₂ = 1, so a = 4 and d = −3.
2. \( S_{20} = \dfrac{20}{2}\left[2 \times 4 + (20-1)(-3)\right] = 10\,[8 - 57] \).
3. S₂₀ = −490.

Teacher's Note:
1. A linear formula for T_n always gives an A.P.; the coefficient of n is the common difference.
2. You can also use \( S_n = \dfrac{n}{2}(T_1 + T_n) \) with T₂₀ = −53.

Answer: (c) 10

Explanation:
1. Here a = 5 and d = 2 − 5 = −3.
2. a + (n − 1)d = −22 ⇒ 5 + (n − 1)(−3) = −22 ⇒ n − 1 = 9.
3. So n = 10: the 10^th term is −22.

Teacher's Note:
1. Set the general term equal to the given value and solve for n.
2. n must come out as a positive integer; otherwise the number is not a term.

Answer: (a) p + q − n

Explanation:
1. a + (p − 1)d = q …(i) and a + (q − 1)d = p …(ii).
2. Subtracting (ii) from (i): (p − q)d = q − p, so d = −1. Then from (i), a = p + q − 1.
3. a_n = (p + q − 1) + (n − 1)(−1) = p + q − n.

Teacher's Note:
1. Two terms of an A.P. fix a and d completely.
2. Check with n = p: p + q − p = q, as given.

Answer: (b) 2m − 1 : 2n − 1

Explanation:
1. \( \dfrac{\frac{m}{2}\left[2a+(m-1)d\right]}{\frac{n}{2}\left[2a+(n-1)d\right]} = \dfrac{m^2}{n^2} \Rightarrow \dfrac{2a+(m-1)d}{2a+(n-1)d} = \dfrac{m}{n} \).
2. Replace m by 2m − 1 and n by 2n − 1: \( \dfrac{2a+2(m-1)d}{2a+2(n-1)d} = \dfrac{2m-1}{2n-1} \).
3. So \( \dfrac{a+(m-1)d}{a+(n-1)d} = \dfrac{a_m}{a_n} = \dfrac{2m-1}{2n-1} \), i.e. 2m − 1 : 2n − 1.

Teacher's Note:
1. S_n ∝ n² happens exactly when d = 2a, as in 1, 3, 5, …
2. Then the terms are a(2n − 1), which gives the ratio at once.

Answer: (b) \( \dfrac{mn+1}{2} \)

Explanation:
1. \( a+(m-1)d = \dfrac{1}{n} \) and \( a+(n-1)d = \dfrac{1}{m} \). Subtracting, \( (m-n)d = \dfrac{1}{n}-\dfrac{1}{m} = \dfrac{m-n}{mn} \), so \( d = \dfrac{1}{mn} \).
2. Then \( a = \dfrac{1}{n} - \dfrac{m-1}{mn} = \dfrac{1}{mn} \).
3. \( S_{mn} = \dfrac{mn}{2}\left[\dfrac{2}{mn} + \dfrac{mn-1}{mn}\right] = \dfrac{mn+1}{2} \).

Teacher's Note:
1. Here a = d, so the k^th term is simply \( \dfrac{k}{mn} \).
2. The sum is then \( \dfrac{1 + 2 + \dots + mn}{mn} \), a neat check.

Answer: (a) 6500 units

Explanation:
1. a + 2d = 6000 and a + 6d = 7000, so 4d = 1000 and d = 250.
2. Then a = 5500.
3. a₅ = 5500 + 4 × 250 = 6500 units.

Teacher's Note:
1. The 5^th term lies exactly midway between the 3^rd and 7^th terms.
2. So it is their average: (6000 + 7000) ÷ 2 = 6500.

Answer: (b) 3500 m

Explanation:
1. The distances of the trees from the well are 10, 15, 20, … m: an A.P. with a = 10, d = 5, n = 25.
2. \( S_{25} = \dfrac{25}{2}\left[2(10) + 24(5)\right] = \dfrac{25}{2} \times 140 = 1750 \) m.
3. He goes to each tree and comes back, so the total distance = 2 × 1750 = 3500 m.

Teacher's Note:
1. Each trip is a round trip, so double the one-way sum.
2. Draw a quick sketch: well, then trees at 10 m, 15 m, 20 m, …

Answer: (b) (n + 1) : n

Explanation:
1. The odd-placed terms a₁, a₃, …, a_2n+1 are n + 1 terms: \( S_1 = \dfrac{n+1}{2}\left[a + a + 2nd\right] = (n+1)(a+nd) \).
2. The even-placed terms a₂, …, a_2n are n terms: \( S_2 = \dfrac{n}{2}\left[(a+d) + (a+(2n-1)d)\right] = n(a+nd) \).
3. So S₁ : S₂ = (n + 1) : n.

Teacher's Note:
1. Both sums equal (number of terms) × (middle term a + nd).
2. There is one more odd-placed term than even-placed terms.

Answer: (d) None of these

Explanation:
1. a₃ + a₇ = 6 and a₃a₇ = 8, so {a₃, a₇} = {2, 4}; with a₇ − a₃ = 4d this gives \( d = \pm\dfrac{1}{2} \).
2. Case I: \( d = \dfrac{1}{2} \), a = 1: \( S_{16} = 8\left[2 + 15 \times \dfrac{1}{2}\right] = 76 \).
3. Case II: \( d = -\dfrac{1}{2} \), a = 5: \( S_{16} = 8\left[10 - \dfrac{15}{2}\right] = 20 \).
4. The sum is 76 or 20; neither is 86 or 90, so the answer is None of these.

Teacher's Note:
1. Two numbers with a given sum and product are the roots of t² − 6t + 8 = 0.
2. Keep both cases: the A.P. may be increasing or decreasing.

Answer: (a) P: \( \dfrac{14m-6}{8m+23} \), Q: n², R: 2

Explanation:
1. (i) \( \dfrac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2} = \dfrac{7n+1}{4n+27} \). Put n = 2m − 1: \( \dfrac{a_1+(m-1)d_1}{a_2+(m-1)d_2} = \dfrac{7(2m-1)+1}{4(2m-1)+27} = \dfrac{14m-6}{8m+23} \).
2. (ii) \( 1 + 3 + 5 + \dots = \dfrac{n}{2}\left[2 + (n-1)2\right] = n^2 \).
3. (iii) \( S_1 = \dfrac{n}{2}(n+1),\ S_2 = n^2,\ S_3 = \dfrac{n}{2}(3n-1) \), so \( \dfrac{S_1+S_3}{S_2} = \dfrac{\frac{n}{2}(4n)}{n^2} = 2 \).
4. So P = \( \dfrac{14m-6}{8m+23} \), Q = n², R = 2.

Teacher's Note:
1. To turn a ratio of sums into a ratio of m^th terms, replace n by 2m − 1.
2. The sum of the first n odd numbers is n² — worth remembering.