KCET · Maths · Sequences and Series
If we insert two numbers between \(\sqrt{2}\) and \(4\) so that the resulting sequence is in G.P, then the inserted numbers in the order are
- A \(8, \sqrt{2}\)
- B \(2, \sqrt{8}\)
- C \(\sqrt{8}, 2\)
- D \(\sqrt{2}, 8\)
Answer & Solution
Correct Answer
(B) \(2, \sqrt{8}\)
Step-by-step Solution
Detailed explanation
Let the geometric progression be \(a, ar, ar^2, ar^3\).
Given \(a = \sqrt{2}\) and \(ar^3 = 4\).
Substituting the value of \(a\), we get:
\(\sqrt{2}r^3 = 4\)
\(r^3 = \dfrac{4}{\sqrt{2}} = 2\sqrt{2} = (\sqrt{2})^3\)
\(r = \sqrt{2}\)
The two inserted numbers are \(ar\) and \(ar^2\).
\(ar = \sqrt{2} \times \sqrt{2} = 2\)
\(ar^2 = \sqrt{2} \times (\sqrt{2})^2 = 2\sqrt{2} = \sqrt{8}\)
The inserted numbers in order are \(2\) and \(\sqrt{8}\).
Answer: \(2, \sqrt{8}\)
Given \(a = \sqrt{2}\) and \(ar^3 = 4\).
Substituting the value of \(a\), we get:
\(\sqrt{2}r^3 = 4\)
\(r^3 = \dfrac{4}{\sqrt{2}} = 2\sqrt{2} = (\sqrt{2})^3\)
\(r = \sqrt{2}\)
The two inserted numbers are \(ar\) and \(ar^2\).
\(ar = \sqrt{2} \times \sqrt{2} = 2\)
\(ar^2 = \sqrt{2} \times (\sqrt{2})^2 = 2\sqrt{2} = \sqrt{8}\)
The inserted numbers in order are \(2\) and \(\sqrt{8}\).
Answer: \(2, \sqrt{8}\)
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