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Question

If a,b,care in HP, then which of the following is true.


A

1(b-a)+1(b-c)=1b

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B

ac(a+c)=b

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C

(b+a)(b-a)+(b+c)(b-c)=1

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D

None of these

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Solution

The correct option is D

None of these


Explanation for the correct option:

Step 1. Given a,b,c are in H.P

b=2ac/(a+c)

Step 2. Check Option (1):

[1/(b-a)]+[1/(b-c)]=1/b

(b-c+b-a)/(b-a)(b-c)=1/b

(2b-c-a)/(b-a)(b-c)=1/b

b(2b-c-a)=(b-a)(b-c)

2b2-bc-ab=b2-ab-bc+ac

b2=ac

Option (1) is wrong

Step 3. Check Option (2);

b=ac(a+c)

Option (2) was wrong

Step 4. Check Option (3);

[(b+a)/(b-a)]+[(b+c)/(b-c)]=1

[(b+a)(b-c)+(b+c)(b-a)]/(b-a)(b-c)=1

(b+a)(b-c)+(b+c)(b-a)=(b-a)(b-c)

b2-bc+ab-ac+b2-ab+bc-ac=b2-ab-bc+ac

b2-2ac=-ab-bc+ac

b2=-ab-bc+3ac

Option (3) was wrong

remaining Option (4) is correct

Hence Option (4) is correct


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