Question

If \(a\), \(b\), \(c\), and \(d\) are in HP and the arithmetic mean of \(ab\), \(bc\), and \(cd\) is 9, then which of the following is the value of \(ad\)?

• A. 3
• B. 9
• C. 12
• D. 4

Hint:

In problems involving HP and AP, first express the variables in terms of their reciprocals and use properties of arithmetic progressions to simplify the problem.

Answer 1

The Correct Option is B

We are given that \(a\), \(b\), \(c\), and \(d\) are in HP. This means the reciprocals of these numbers are in AP. Therefore, we can write: 
\[ \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d} \] are in AP, implying: 
\[ \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} = \frac{1}{d} - \frac{1}{c} \] Let this common difference be \( k \). Thus, we have the following relationships: 
\[ \frac{1}{b} - \frac{1}{a} = k \Rightarrow b = \frac{a}{a+k} \] \[ \frac{1}{c} - \frac{1}{b} = k \Rightarrow c = \frac{b}{b+k} \] \[ \frac{1}{d} - \frac{1}{c} = k \Rightarrow d = \frac{c}{c+k} \] Next, we are told that the arithmetic mean of \(ab\), \(bc\), and \(cd\) is 9. Therefore: 
\[ \frac{ab + bc + cd}{3} = 9 \] This gives: \[ ab + bc + cd = 27 \] Substitute the expressions for \(b\), \(c\), and \(d\) in terms of \(a\) and \(k\). After solving for \(a\) and \(d\), we find: 
\[ ad = 9 \] Thus, the value of \(ad\) is \( \boxed{9} \).