If a , b , c are in H . P . show that
ad > bc

Open in App

Solution

a , b , c , d are in H . P .
b is the H . M of a ,and c are there A . M is $\frac{a + b}{2}$ and G . M is $\sqrt{a} c$
c is the H . m of b and d are their A . M is $\frac{b + d}{2}$ and G . M is $\sqrt{b d}$
Also the A > G > M where A , G , H are respectively means
$∴ \frac{a + c}{2} > b a n d \frac{b + d}{2} > c ∵ A > H$
Add $, \frac{a + c}{2} + \frac{b + d}{2} > B + c o r a + b + c + d > 2 a b + 2 c O r a + b > b + c .$ $T h i s i s (b) \sqrt{a} c > b a n d \sqrt{b} d > c ∵ G > H$
Multipy $∴ \sqrt{A} c \sqrt{b} d > b c o r a c b d > b^{2} c^{2} o r a d > b c$ This is (a)

Suggest Corrections

Similar questions

D

B C

A B C

A D > A C

A B > A D

In △ ABC, seg DE ∥ side BC . If 2 A (△ ADE) = A (□ DBCE), f ind AB : AD and show that BC = \sqrt{3} \times DE .

∠ B < ∠ A

∠ C < ∠ D

A D < B C

Join BYJU'S Learning Program