Exams › JEE Advanced › Maths

The n-th term of a sequence is Tₙ = (a - 3)*n³ + (b - 4)*n² + (a + b)*n - 4 for all natural numbers n. If this sequence is an arithmetic progression, which of the following is/are true?

1. a² + b² = 25
2. The common difference of the AP is 7
3. The common difference of the AP is 9
4. The sum of the first 10 terms of the AP is 345

Correct answer: a² + b² = 25

Solution

For an arithmetic progression, Tₙ must be linear in n. So coefficients of n³ and n² must vanish: a - 3 = 0 giving a = 3, and b - 4 = 0 giving b = 4. Then Tₙ = 0 + 0 + (3+4)*n - 4 = 7n - 4. First term T₁ = 7 - 4 = 3, common difference d = 7. Sum of 10 terms S₁₀ = (10/2)*(2*3 + 9*7) = 5*(6 + 63) = 5*69 = 345. Also a² + b² = 9 + 16 = 25.

Related JEE Advanced Maths questions

• If a, b, and c are in harmonic progression, then e raised to the power of -a, e raised to the power of -b, and e raised to the power of -c will be in which progression?
• If x, y, and z represent the pᵗʰ, qᵗʰ, and rᵗʰ terms of both an arithmetic progression and a geometric progression, what is the value of (xʳ)(yᵖ)(zᵠ)?
• Let ϕ(x) represent a quadratic polynomial. Given that ϕ(1) equals ϕ(−1) and the terms a₁, a₂, a₃ form an arithmetic progression, then the values ϕ(a₁), ϕ(a₂), ϕ(a₃) will be in which sequence?
• Let Sₙ = Σ (k+1)/2 * k². Then Sₙ can take value(s)
• If a, b, and c are positive integers such that b is divisible by a, and they form a geometric sequence, while their arithmetic mean equals b + 2, what is the value of (a² + a - 14)/(a + 1)?
• Let bᵢ > 1 for i = 1, 2,..., 101. Assume that logₑb₁, logₑb₂,..., logₑb₁₀₁ form an arithmetic sequence with a common difference of log₂. Also, let a₁, a₂,..., a₁₀₁ form an arithmetic sequence where a₁ = b₁ and a₅₁ = b₅₁. If t represents the sum b₁ + b₂ +... + b₅₁ and s represents the sum a₁ + a₂ +... + a₅₁, then which of the following is true?

⚔️ Practice JEE Advanced Maths free + battle 1v1 →