Mathematics · 2025
JEE Main · 22 January 2025, Shift 1 · Q5
If Σ_r=1^n T_r=((2 n-1)(2 n+1)(2 n+3)(2 n+5))/64, then lim_n → ∞ Σ_r=1^n(1/T_r) is equal to:
If $\displaystyle \sum_{r=1}^n T_r=\frac{(2 n-1)(2 n+1)(2 n+3)(2 n+5)}{64}$, then $\displaystyle \lim _{n \rightarrow \infty} \sum_{r=1}^n\left(\frac{1}{T_r}\right)$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle \frac{2}{3}$
More from Progressions
- Let a_1, a_2, a_3, a_4 be an A.P. of four terms such that each term of the A.P. and its common difference l are integers. If a_1+a_2+a_3+a_4=48 and…2026
- 6/(3^26)+(10 · 1)/(3^25)+(10 · 2)/(3^24)+(10 · 2^2)/(3^23)+…+(10 · 2^24)/3 is equal to:2026
- Let a_1, a_2, a_3, …. be an A.P. and g_1=a_1, g_2, g_3, …. be an increasing G.P. If a_1=a_2+g_2=1 and a_3+g_3=4, then a_10+g_5 is equal to:2026
- Let A={1,6,11,16, …} and B={9,16,23,30, …} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A ∪ B) is2025
- Let <a_n> be a sequence such that a_0=0, a_1=1/2 and 2 a_n+2=5 a_n+1-3 a_n, n=0,1,2,3, …. Then Σ_k=1^100 a_k is equal to2025
- Let A be the set of first 101 terms of an A.P., whose first term is 1 and the common difference is 5 and let B be the set of first 71 terms of an…2026
- (1/3+4/7)+(1/3^2+1/3 × 4/7+4^2/7^2)+(1/3^3+1/3^2 × 4/7+1/3 × 4^2/7^2+4^3/7^3)+… upto infinite terms, is equal to2026
- The value of lim_n → ∞(Σ_k=1^n (k^3+6 k^2+11 k+5)/((k+3)!)) is:2025
JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.