JEE Main 2015 · 4 Apr · Shift 1 · Mathematics Q68

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MathematicsSingle correct+4 / -1Question 68 of 90

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.

4l^2mn

4lm^2n

4lmn^2

4l^2m^2n^2

Answer and worked solution

Official answer: B

With ratio $r$ , $r^4=\frac nl$ and $G_k=lr^k$ :
$G_1^4+2G_2^4+G_3^4=l^4r^4(1+r^4)^2=l^3n\left(\frac{l+n}l\right)^2=ln(l+n)^2=4lm^2n.$

Topics:Geometric progression Arithmetic progression

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JEE Main 2015 · 4 Apr · Shift 1, Mathematics question 68