JEE Mains · Maths · STD 11 - 8. sequence and series
For the two positive numbers \(a , b\), if \(a , b\) and \(\frac{1}{18}\) are in a geometric progression, while \(\frac{1}{ a }, 10\) and \(\frac{1}{ b }\) are in an arithmetic progression, then, \(16 a+12 b\) is equal to \(.........\).
- A \(3\)
- B \(2\)
- C \(1\)
- D \(0\)
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
\(a , b , \frac{1}{18} \rightarrow GP\) \(\frac{ a }{18}= b ^2\) \(\frac{1}{ a }, 10, \frac{1}{ b } \rightarrow AP\) \(\frac{1}{ a }+\frac{1}{ b }=20\) \(\Rightarrow a + b =20 ab , \text { from eq. (i) } ; \text { we get }\) \(\Rightarrow 18 b ^2+ b =360 b ^3\)…
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