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

माना \(a_{1}, a_{2}, a_{3}, \ldots ., a_{49}\) एक समांतर श्रेढ़ी में ऐसे है कि \(\sum_{k=0}^{12} a_{4 k+1}=416\) तथा \(a_{9}+a_{43}=66\) है। यदि \(a_{1}^{2}+a_{2}^{2}+\ldots . .+a_{17}^{2}=140\, m\) है, तो \(m\) बराबर है

  1. A \(68\)
  2. B \(34\)
  3. C \(33\)
  4. D \(66\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(34\)

Step-by-step Solution

Detailed explanation

(2) \(\because\) \(\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416 \Rightarrow \frac{{13}}{2}\left[ {2{a_1} + 48d} \right] = 416\) \( \Rightarrow {a_1} + 24d = 32\,\,\,\,\,\,\,\,..........\left( 1 \right)\) Now,…
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