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

ધારો કે \(a_1, a_2, a_3 \ldots\) સમાંતર શ્રેણીમાં છે, જેથી \(\sum_{\mathrm{k}=1}^{12} \mathrm{a}_{2 \mathrm{k}-1}=-\frac{72}{5} \mathrm{a}_1, \mathrm{a}_1 \neq 0\). જો \(\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{k}}=0\), તો n = __________

  1. A 11
  2. B 10
  3. C 18
  4. D 17
Verified Solution

Answer & Solution

Correct Answer

(A) 11

Step-by-step Solution

Detailed explanation

Let \(a_1=a\), common difference \(=d\) \(a_1+a_3+a_5+\ldots \ldots+a_{23}=-\frac{72}{5} a\) \(\frac{12}{2}[2 a+11 \times 2 d]=-\frac{72}{5} a\) \(12 a+132 d=-\frac{72}{5} a\) \(132 a+132 \times 5 d=0\) \(\mathrm{a}=-5 \mathrm{~d}\)…
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