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JEE Mains · Maths · STD 11 - 8. sequence and series

माना \(<\mathrm{a}_{\mathrm{n}}>\) एक अनुक्रम है जिसके लिए \(a_1+a_2+\ldots+a_n=\frac{n^2+3 n}{(n+1)(n+2)}\) है। यदि \(28 \sum_{\mathrm{k}=1}^{10} \frac{1}{\mathrm{a}_{\mathrm{k}}}=\mathrm{p}_1 \mathrm{p}_2 \mathrm{p}_3 \ldots . \mathrm{p}_{\mathrm{m}}\), जहाँ \(\mathrm{p}_1, \mathrm{p}_2, \ldots . \mathrm{p}_{\mathrm{m}}\) प्रथम \(\mathrm{m}\) अभाज्य संख्याएँ है, तो \(\mathrm{m}\) बराबर है :

  1. A \(7\)
  2. B \(6\)
  3. C \(5\)
  4. D \(8\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(6\)

Step-by-step Solution

Detailed explanation

\(a_n=S_n-S_{n-1}=\frac{ n ^2+3 n }{( n +1)( n +2)}-\frac{( n -1)( n +2)}{ n ( n +1)}\) \(\Rightarrow a _{ n }=\frac{4}{ n ( n +1)( n +2)}\) \(\Rightarrow 28 \sum \limits_{ k =1}^{10} \frac{1}{ a _{ k }}=28 \sum \limits_{ k =1}^{10} \frac{ k ( k +1)( k +2)}{4}\)…
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