Latest Elitmus Aptitude Question SOLUTION: if a is a arithmetic mean and b, c be the geometric mean between any two positive numbers, then (b^3+c^3)/abc equals....?

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(#M40038209) ELITMUS Category question Keep an EYE

if a is a arithmetic mean and b, c be the geometric mean between any two positive numbers, then (b^3+c^3)/abc equals....?

Asked In Elitmus ashish (9 years ago)

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RAKESH  (9 years ago)

ans: 2

if x,y be two no.s then

x,a,y are in AP => 2a=(x+y)

x,b,c,y are in GP => b=xr , c = xr^2 where r^3 = y/x and r is cr.

(b^3+c^3)= x^3*r^3 + x^3*r^6 = x^3*r^3(1+r^3) = xy(x+y)

bc = xr * xr^2 = x^2 *r^3 = xy

(b^3+c^3)/abc = xy(x+y) / (x+y)/2 * xy = 2

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balagopalakrishna  (9 years ago)

ie taking x and y are two numbers
so x,a,y are in ap and AM => 2a=x+y
and x,b,c,y are in gp

and by taking two possibilities x,b,c are in gp and
b,c,y are in gp
so GM for above two is
b^2=xc and c^2= by

but we need to find

(b^3+c^3)/abc dividing both terms by abc

(b^2/ac)+(c^2/ab)
we know b^2 and c^2

xc/ac + by/ab

ie x+y/a

subtituting a=x+y/2

so we get 2
ans ==2

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Dipanjan Dasgupta  (9 years ago)

how can there be 2 geometric mean values???

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T.VIGNESWARA REDDY  (9 years ago)

i think a os A.M b is G.M c is H.M so a.c=b*b so ans is 1+(c/b)pow3
a=(x+y)/2 b=sqrt(xy) c is 2xy/x+y

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T.VIGNESWARA REDDY  (9 years ago)

rakesh is right two g.m possible b=sqrt(xc) c=sqrt(by)

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