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

माना \(\left\{a_{\mathrm{k}}\right\}\) तथा \(\left\{\mathrm{b}_{\mathrm{k}}\right\}, \mathrm{k} \in \mathbb{N}\), दो G.P. है, जिनके सार्व अनुपात क्रमशः \(r_1\) तथा \(r_2\) है और \(a_1=b_1=4\), \(\mathrm{r}_1<\mathrm{r}_2\) है। माना \(\mathrm{c}_{\mathrm{k}}=\mathrm{a}_{\mathrm{k}}+\mathrm{b}_{\mathrm{k}}, \mathrm{k} \in \mathbb{N}\) है। यदि \(\mathrm{c}_2=5\) तथा \(\mathrm{c}_3=\frac{13}{4}\) है तो \(\sum_{\mathrm{k}=1}^{\infty} \mathrm{c}_{\mathrm{k}}-\left(12 \mathrm{a}_6+8 \mathrm{~b}_4\right)\) बराबर है________. 

  1. A \(9\)
  2. B \(18\)
  3. C \(20\)
  4. D \(22\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(9\)

Step-by-step Solution

Detailed explanation

Given that \(c_k=a_k+b_k\) and also \(a _2=4 r _1\) \(\quad a _3=4 r _1{ }^2\) \(b _2=4 r _2\) \(\quad b _3=4 r _2{ }^2\) Now \(c_2=a_2+b_2=5\) and \(c_3=a_3+b_3=\frac{13}{4}\) \(\Rightarrow r_1+r_2=\frac{5}{4}\) and \(r_1^2+r_2^2=\frac{13}{16}\) Hence \(r_1 r_2=\frac{3}{8}\)…
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