Aspire Faculty ID #11517 · Topic: NIMCET 2023 · Just now
NIMCET 2023

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

Solution

$a, b, c, d$ are in H.P. 

$\Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in A.P.
 
$\Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c}$ and $\frac{2}{c} = \frac{1}{b} + \frac{1}{d}$


Given A.M. of $ab, bc, cd$ is $9$: 

$\frac{ab + bc + cd}{3} = 9$ 

$\Rightarrow ab + bc + cd = 27$

Now multiply: $\frac{2}{b} = \frac{1}{a} + \frac{1}{c} $

$\Rightarrow 2ac = b(a + c)$

$\frac{2}{c} = \frac{1}{b} + \frac{1}{d} \Rightarrow 2bd = c(b + d)$

Multiply both: $4abcd = bc(a + c)(b + d)$

Cancel $bc$: 

$4ad = (a + c)(b + d)$

Expand RHS: 

$4ad = ab + ad + bc + cd$

$\Rightarrow 3ad = ab + bc + cd$
 
But $ab + bc + cd = 27$: 

$\Rightarrow 3ad = 27$ 

$\Rightarrow ad = 9$

$\boxed{9}$
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