JEE Mains · Maths · STD 11 - 8. sequence and series
If \(m\) is the \(A.M\) of two distinct real numbers \( l\) and \(n (l,n>1) \) and \(G_1, G_2\) and \(G_3\) are three geometric means between \(l\) and \(n\) then \(G_1^4 + 2G_2^4 + G_3^4\) equals :
- A \(4{l^2}{m^2}{n^2}\)
- B \(4{l^2}mn\)
- C \(4l{m^2}n\)
- D \(4lm{n^2}\)
Answer & Solution
Correct Answer
(C) \(4l{m^2}n\)
Step-by-step Solution
Detailed explanation
\(m=\frac{l+n}{2}\) \(\Rightarrow 2 m=l+n\) \(G_{1}, G_{2}, G_{3}\) \(l, G_{1}, G_{2}, G_{3},n\) are in \(GP\) let \(d\) be the common ration \(G_{1}=l d\) \(G_{2}=l d^{2}\) \(G_{3}=l d^{3}\) \(n=l d^{4}\)…
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