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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 9x2−10x+1=0if 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)=10if ,x,y,z,b are in HP then their reciprocal will be in AP1a+1x+1y+1z+1b=52(1a+1b)(1a+1b+53)×2=5×(1a+1b)103=3(a+b)abab=9 (from eq (1) putting value of (a+b))if roots are 1a,1bsum of roots =a+bab=109product of roots =1ab=19quadratic eq:x2−(sum of the roots)x+product of the roots=0x2−109x+19=09x2−10x+1=0

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