Sequences and Series: JEE Main PYQ, 2020–2025

Answer: (A)

For an AP, the sum of the first n terms is Sₙ = (n/2)[2a + (n − 1)d]. Here a = 5, d = 3 and n = 20, so S₂₀ = 10 × [10 + 19 × 3] = 10 × 67 = 670. As a check, the 20th term is 5 + 19 × 3 = 62, and (20/2)(5 + 62) = 670.

Answer: (A)

Set (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) = (a + b)/2. Cross-multiplying gives 2aⁿ⁺¹ + 2bⁿ⁺¹ = aⁿ⁺¹ + abⁿ + aⁿb + bⁿ⁺¹. This simplifies to aⁿ⁺¹ + bⁿ⁺¹ − abⁿ − aⁿb = 0, which factors as (aⁿ − bⁿ)(a − b) = 0. Since a ≠ b, we need aⁿ = bⁿ, so n = 0. Check: n = −1/2 gives √(ab), and n = −1 gives the harmonic mean.

Answer: (A)

A number common to P and Q must satisfy x ≡ 3 (mod 4) and x ≡ 2 (mod 7). Testing 3, 7, 11, 15, 19, 23, … in P, the first one that is ≡ 2 (mod 7) is 23, since 23 − 2 = 21. Common terms of two APs form an AP whose common difference is the lcm of the two differences, so here it is 28. The common terms are therefore 23, 51, 79, … To lie in both progressions they must be at most min(407, 709) = 407. From 23 + 28k ≤ 407 we get k ≤ 13.71, so k = 0, 1, …, 13, which is 14 terms. The last term is 23 + 28 × 13 = 387. The sum is 14 × (23 + 387)/2 = 7 × 410 = 2870.

Answer: (A)

The AM–GM inequality says (a+b+c)/3 ≥ (abc)^(1/3) for positive numbers, with equality only if a = b = c. Testing 1, 2, 3 (AP) and 1, 2, 4 (GP) shows AM > GM in both cases, so only equal numbers give equality.

Answer: (A)

Write 0.4373737… = 0.4 + 0.0373737…. The second part is the infinite GP 0.037 + 0.00037 + … with first term 37/1000 and ratio 1/100, so its sum is (37/1000)/(1 − 1/100) = 37/990. Total = 4/10 + 37/990 = 396/990 + 37/990 = 433/990. 990 = 2·3²·5·11 and 433 is prime, so the fraction is in lowest terms. Then p + q = 1423.

Answer: (A)

Take x > 0 first. By A.M. ≥ G.M., 3^x + 3^(1/x) ≥ 2√(3^(x+1/x)) = 2·3^((x+1/x)/2). Since x + 1/x ≥ 2 for x > 0, this is at least 2·3 = 6. Equality needs both 3^x = 3^(1/x) and x + 1/x = 2, so x = 1. Now take x < 0: then x and 1/x are both negative, so 3^x < 1 and 3^(1/x) < 1, and the sum is less than 2, so there is no solution. The point x = 0 is excluded. So there is exactly one solution, x = 1.

Answer: (A)

Inserting three means gives a GP of 5 terms, with 3 as the first term and 48 as the fifth. So 3r^4 = 48, r^4 = 16 and r = 2 (positive). The means are 6, 12 and 24. Their sum is 6 + 12 + 24 = 42.

Answer: (A)

Let the odd-placed sum be S_odd = a + ar² + … + ar²ⁿ⁻². The even-placed terms are ar, ar³, …, each r times the corresponding odd-placed term, so S_even = r·S_odd. The total is S_odd(1 + r) = 5 S_odd, hence 1 + r = 5 and r = 4.

Answer: (A)

Let t = √(a/b). Then (a + b)/(2√(ab)) = (t² + 1)/(2t) = 5/4, so 2t² − 5t + 2 = 0 and t = 2 (since a > b). So a/b = 4. Inserting 3 GMs gives 5 terms a, ar, ar², ar³, ar⁴ = b. Thus r⁴ = b/a = 1/4 and the positive ratio is r = 1/√2.

Answer: (A)

For positive numbers, a + b ≥ 2√(ab), b + c ≥ 2√(bc) and c + a ≥ 2√(ca). Multiplying gives (a+b)(b+c)(c+a) ≥ 8√(a²b²c²) = 8abc. So the ratio is at least 8, and a = b = c gives exactly 8.