Question

If \(p(\frac{1}{q}+\frac{1}{r}),q(\frac{1}{r}+\frac{1}{p}),r(\frac{1}{p}+\frac{1}{q})\) are in A.P., then p, q, r

• A. are in A.P.
• B. are not in A.P.
• C. are not in G.P.
• D. are in G.P.

Hint:

When dealing with sequences or series, remember that for terms to be in Arithmetic Progression (A.P.), the difference between consecutive terms must be constant. In problems involving algebraic expressions, you can use this property by equating the differences between the terms and simplifying the resulting equations. In this case, simplifying the expressions involving p, q, r will show that they form an A.P. when these variables themselves are in A.P.

Answer 1

The Correct Option is A

The correct answer is (A): are in A.P.

Answer 2

The correct answer is: (A) are in A.P.

The given expressions are:

\( p\left(\frac{1}{q} + \frac{1}{r}\right), q\left(\frac{1}{r} + \frac{1}{p}\right), r\left(\frac{1}{p} + \frac{1}{q}\right) \)

These expressions are stated to be in Arithmetic Progression (A.P.). In an A.P., the difference between consecutive terms is constant. Therefore, the second term minus the first term should equal the third term minus the second term:

Let’s calculate the differences:

• Second term - First term: \( q\left(\frac{1}{r} + \frac{1}{p}\right) - p\left(\frac{1}{q} + \frac{1}{r}\right) \)
• Third term - Second term: \( r\left(\frac{1}{p} + \frac{1}{q}\right) - q\left(\frac{1}{r} + \frac{1}{p}\right) \)

After simplifying these expressions, we find that the condition for the terms to be in A.P. holds true when p, q, and r are in arithmetic progression (A.P.).

Thus, the correct answer is (A) are in A.P..