If m is the A-M- of two distinct real numbers I and n1- n-1 and G1- G2 and G3 are three geometric means between 1 and n- then G14-2 G24-G34 equals-

The correct option is D 4lm^2nm = l+n/2 and common ratio of G.P. = r = ( n/l )^1/4 ∴ G_1 = l^3/4 n^1/4, G_2 = l^1/2n^1/2, G_3 = l^1/4 n^3/4 G^4_1 + 2G^4_2 + ...

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Byju's Answer

Standard XII

Mathematics

Insertion of GP's between 2 Numbers

If m is the A...

Question

If m is the A.M. of two distinct real numbers I and n(l, n > 1) and $G_{1}, G_{2}$ and $G_{3}$ are three geometric means between l and n, then $G_{1}^{4} + 2 G_{2}^{4} + G_{3}^{4}$ equals:

4 l m n^{2}

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4 l^{2} m^{2} n^{2}

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4 l^{2} m n

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4 l m^{2} n

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Solution

The correct option is D $4 l m^{2} n$
$m = \frac{l + n}{2}$ and common ratio of G.P. $= r = {(\frac{n}{l})}^{\frac{1}{4}} ∴ G_{1} = l^{3 / 4} n^{1 / 4}, G_{2} = l^{1 / 2} n^{1 / 2}, G_{3} = l^{1 / 4} n^{3 / 4} G_{1}^{4} + 2 G_{2}^{4} + G_{3}^{4} = l^{3} n + 2 l^{2} n^{2} + l n^{3} = l n (l + n)^{2} = l n \times {2 m}^{2} = 4 l m^{2} n$

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If m is the A.M. of two distinct real numbers I and n(l, n > 1) and G1,G2 and G3 are three geometric means between l and n, then G41+2G42+G43 equals:

The correct option is D 4lm2nm=l+n2 and common ratio of G.P. =r=(nl)14∴G1=l3/4 n1/4,G2=l1/2n1/2,G3=l1/4 n3/4G41+2G42+G43=l3n+2l2n2+ln3=ln(l+n)2=ln×2m2=4lm2n

If m is the A.M. of two distinct real numbers I and n(l, n > 1) and $G_{1}, G_{2}$ and $G_{3}$ are three geometric means between l and n, then $G_{1}^{4} + 2 G_{2}^{4} + G_{3}^{4}$ equals: