Exams › JEE Advanced › Maths
Find the sum of the first 20 terms of the series: 1 + (1+2) + (1+2+3) +...
- 1410
- 210
- 1305
- 1540
Correct answer: 1540
Solution
The nth term is n*(n+1)/2. Summing from n=1 to 20: (1/2)*(sum of n² from 1 to 20 + sum of n from 1 to 20) = (1/2)*(2870 + 210) = (1/2)*3080 = 1540.
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 →