Sequences and Series: JEE Advanced PYQ, 2007–2026

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)

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)

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)

In a GP, aₘ · aₙ depends only on m + n, since aₘaₙ = a²r^(m+n−2). Because 1 + 10 = 2 + 9 = 3 + 8 = 4 + 7 = 5 + 6 = 11, each of these pairs has the same product as a₄a₇ = 9. The product of all ten terms is 9⁵ = 3¹⁰ = 59049.

Answer: (A)

Let the roots be b/r, b, br. For x³ − 7x² + px − 8 = 0, the product of the roots is +8, so b³ = 8 and b = 2. Since b is a root, substitute x = 2: 8 − 28 + 2p − 8 = 0, so p = 14. Check: the sum of the roots is 2(1/r + 1 + r) = 7, so r + 1/r = 5/2 and r = 2 or 1/2. The roots are 1, 2, 4, and the pairwise sum is 2 + 8 + 4 = 14. ✓

Answer: (A)

The first progression has terms 3k + 1, so they leave remainder 1 on division by 3. The second progression has terms 5m − 2, so they leave remainder 3 on division by 5. A common term n must satisfy n ≡ 1 (mod 3) and n ≡ 3 (mod 5). The first such number is 13, and the others differ from it by multiples of 15. So the common terms form an AP with first term 13 and common difference 15. The 50th term of the first progression is 4 + 49·3 = 151, and the 40th term of the second is 3 + 39·5 = 198. A common term must be at most min(151, 198) = 151. From 13 + 15k ≤ 151 we get k ≤ 9.2, so k = 0, 1, …, 9, which is 10 terms. The last is 13 + 135 = 148. The sum is 10 × (13 + 148)/2 = 805.

Answer: (A)

The constraint involves x², so choose terms whose product is constant: x/2, x/2 and y. Their product is x²y/4 = 8, which is constant. AM–GM gives (x/2 + x/2 + y)/3 ≥ 8^(1/3) = 2, so x + y ≥ 6. Equality needs x/2 = y, so x = 2y. Then 4y²·y = 32, so y = 2 and x = 4, and the sum is 6 (attained).