Let a, b, c be in A.P. Consider the following statements: 1. 1/ab,

1.	Let a, b, c be in A.P. Consider the following statements: 1. 1/ab, 1/ca and 1/bc are in AP. 2. 1/(b+c) ,1/(c+a) and 1/(a+b) are in A.P. Which of the statements given above is/are correct?1. 1 only 2. 2 only 3. Both 1 and 24. Neither 1 nor 2
Answer» Correct Answer - Option 1 : 1 only Concept: If a,b, c are in AP, then 2b = a + c or b - a = c - b Calculation: Let, 1/ab, 1/ca and 1/bc are in AP ⇒\(\rm \frac{1}{ca}\rm -\frac{1}{ab}=\rm \frac{1}{bc}-\rm \frac{1}{ca}\) \(\rm \frac{1}{a}( \frac{1}{a}- \frac{1}{a})= \frac{1}{a}( \frac{1}{a}- \frac{1}{a})\\ ⇒ \frac{b-c}{abc}= \frac{a-b}{abc}\) ⇒ b-c = a-b ⇒2b = a+c ⇒ a,b, c are in AP, which is true. Now, let 1/(b+c) ,1/(c+a) and 1/(a+b) are in AP \(\rm \frac{2}{c+a}=\frac{1}{b+c}+\frac{1}{a+b}\) ⇒ 2(a+b)(b+c) = (a+b+b+c)(c+a) ⇒2(ab + ac + b² + bc) = (ac + a² + 2bc + 2ab + c² + ca) ⇒2ab + 2ac + 2b² + 2bc = ac + a² + 2bc + 2ab + ac + c² ⇒ 2b² = a² + c² a, b, c are not in AP, so this is not true. Hence, option (1) is correct.

Let a, b, c be in A.P. Consider the following statements: 1. 1/ab, 1/ca and 1/bc are in AP. 2. 1/(b+c) ,1/(c+a) and 1/(a+b) are in A.P. Which of the statements given above is/are correct?1. 1 only 2. 2 only 3. Both 1 and 24. Neither 1 nor 2

Answer» Correct Answer - Option 1 : 1 only

Concept:

If a,b, c are in AP, then 2b = a + c or b - a = c - b

Calculation:

Let, 1/ab, 1/ca and 1/bc are in AP

⇒\(\rm \frac{1}{ca}\rm -\frac{1}{ab}=\rm \frac{1}{bc}-\rm \frac{1}{ca}\)

\(\rm \frac{1}{a}( \frac{1}{a}- \frac{1}{a})= \frac{1}{a}( \frac{1}{a}- \frac{1}{a})\\ ⇒ \frac{b-c}{abc}= \frac{a-b}{abc}\)

⇒ b-c = a-b

⇒2b = a+c

⇒ a,b, c are in AP, which is true.

Now, let 1/(b+c) ,1/(c+a) and 1/(a+b) are in AP

\(\rm \frac{2}{c+a}=\frac{1}{b+c}+\frac{1}{a+b}\)

⇒ 2(a+b)(b+c) = (a+b+b+c)(c+a)

⇒2(ab + ac + b² + bc) = (ac + a² + 2bc + 2ab + c² + ca)

⇒2ab + 2ac + 2b² + 2bc = ac + a² + 2bc + 2ab + ac + c²

⇒ 2b² = a² + c²

a, b, c are not in AP, so this is not true.

Hence, option (1) is correct.