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Question

x+y+z=15 when a,x,y,z,b are in AP and 1x+1y+1z=53 when a, x, y, z, b are in HP then the quadratic equation whose roots are 1aand1b is

A
x210x+9=0
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B
x2+10x9=0
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C
9x210x+1=0
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D
9x2+10x+1=0
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Solution

The correct option is C 9x210x+1=0
if a,x,y,z,b are in AP.
a+x+y+z+b=52(a+b)
(a+b+15)×2=5(a+b)
3(a+b)=30
(a+b)=10
if ,x,y,z,b are in HP then their reciprocal will be in AP

1a+1x+1y+1z+1b=52(1a+1b)
(1a+1b+53)×2=5×(1a+1b)
103=3(a+b)ab
ab=9 (from eq (1) putting value of (a+b))


if roots are 1a,1b
sum of roots =a+bab
=109
product of roots =1ab
=19

quadratic eq:
x2(sum of the roots)x+product of the roots=0
x2109x+19=0
9x210x+1=0

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