JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(\frac{1}{16}, a\) and \(b\) be in \(G.P.\) and \(\frac{1}{ a }, \frac{1}{ b }, 6\) be in \(A.P.,\) where \(a , b >0\). Then \(72( a + b )\) is equal to ...... .
- A \(12\)
- B \(18\)
- C \(14\)
- D \(21\)
Answer & Solution
Correct Answer
(C) \(14\)
Step-by-step Solution
Detailed explanation
\(a^{2}=\frac{b}{16} \Rightarrow \frac{1}{b}=\frac{1}{16 a^{2}}\) \(\frac{2}{b}=\frac{1}{a}+6\) \(\frac{1}{8 a^{2}}=\frac{1}{a}+6\) \(\frac{1}{a^{2}}-\frac{8}{a}-48=0\) \(\frac{1}{a}=12,-4 \Rightarrow a=\frac{1}{12},-\frac{1}{4}\) \(a=\frac{1}{12}, a>0\)…
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