JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(\left\{a_k\right\}\) and \(\left\{b_k\right\}, k \in N\), be two G.P.s with common ratio \(r_1\) and \(r_2\) respectively such that \(a_1=b_1=4\) and \(r_1 < r_2\). Let \(c_k=a_k+k, \in N\). If \(c_2=5\) and \(c_3=13 / 4\) then \(\sum \limits_{k=1}^{\infty} c_k - \left(12 a _6+8 b _4\right)\) is equal to
- A \(9\)
- B \(18\)
- C \(20\)
- D \(22\)
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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