Question

If a,b,c,d are in continued proportion,prove that:
(in the above image)

ii)(a²-b²) (c²-d²)=(b²-c²)²

Answer 1

Hope it'd help you..... ..

Answer 2

(a² - b²)(c² - d²) = (b² - c²)² Proved.

Step-by-step explanation:

The numbers a, b, c, and d are in continued proportion.

So, $\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k$ (Say)

So, a  = kb, b = kc and c = kd

⇒ a = k²c = k³d

and, b = kc = k²d

Then, we have to prove that (a² - b²)(c² - d²) = (b² - c²)²

Now, Left hand side

= (a² - b²)(c² - d²)

= (a + b)(a - b)(c + d)(c - d)

= (k³d + k²d)(k³d - k²d)(kd + d)(kd - d)

= $d^{4}k^{4}(k + 1)^{2} (k - 1)^{2}$

Now, Right hand side

= (b² - c²)²

= (b + c)²(b - c)²

= (k²d + kd)² (k²d - kd)²

= $d^{4}k^{4}(k + 1)^{2} (k - 1)^{2}$

Hence, proved that LHS = RHS.