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Question

Let p and q be roots of the equation x2−2x+A=0 and let r and s be the roots of the equation x2−18x+B=0. If p<q< r<s are in arithmetic progression, then the values of A and B are

A
3,77
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B
3,77
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C
3,77
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D
3,77
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Solution

The correct option is D 3,77
Given, Let p and q be roots of the equation x22x+A=0 and let r and s be the roots of the equation x218x+B=0.

Let the four numbers in A.P. be p=a3d,q=ad,r=a+ d,s=a+3d.

Therefore,

p+q=2,r+s=18

Given that pq=A,rs=B.

p+q+r+s=4a=20

a=5

Now,

p+q=2104d=2

r+s=1810+4d=18

d=2

Hence, the numbers are 1,3,7,11

pq=A=3,rs=B=77

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