Math miscellaneousHard
Question
If m is the A.M. of two distinct real numbers l and n(l, n > 1) and G1, G2 and G3 are three geometric means between l and n, then G41 + 2G42 + G43 equals.
Options
A.4 lmn2
B.4 l2m2n2
C.4 l2mn
D.4 lm2n
Solution
l, G1, G2, G3, n in G.P.
Let r be the common ratio ⇒ r4 =
Here G14 + 2G24 + G34 = (lr)4 + 2(lr2)4 + (lr3)4
= nl[l2 + 2ln + n2] = nl(l + n)2 = nl4m2
= 4m2nl (∵ 2m = n + l)
Let r be the common ratio ⇒ r4 =
Here G14 + 2G24 + G34 = (lr)4 + 2(lr2)4 + (lr3)4
= nl[l2 + 2ln + n2] = nl(l + n)2 = nl4m2
= 4m2nl (∵ 2m = n + l)
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