JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(a_1, a_2, \ldots \ldots, a_n\) be in A.P. If \(a_5=2 a_7\) and \(a_{11}=18\), then \(12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots . \cdot \frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)\) is equal to \(..........\).
- A \(8\)
- B \(6\)
- C \(3\)
- D \(12\)
Answer & Solution
Correct Answer
(A) \(8\)
Step-by-step Solution
Detailed explanation
\(a_5=2a_7 \implies a+4d=2(a+6d) \implies a=-8d\) \(a_{11}=18 \implies a+10d=18 \implies -8d+10d=18 \implies 2d=18 \implies d=9\) \(a=-8(9)=-72\) \(\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{a_{k+1}-a_k} = \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}\)…
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