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

ધારો કે \({a_1},{a_2},\;.\;.\;.\;.,{a_{49}}\) સમાંતર શ્રેણીમાં છે તથા \(\mathop \sum \limits_{k = 0}^{12} {a_{4k + 1}} = 416\) અને \({a_9} + {a_{43}} = 66\). જો \(a_1^2 + a_2^2 + \ldots + a_{17}^2 = 140m,\) તો \(m = \;\;..\;.\;.\;.\;\)

  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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