If a, b, c are in A.P., then $(a - c)^{2} = 4 (b^{2} - a c)$ .

True

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

False

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A True
We have,
$a, b, c$ are in $A . P$

Therefore,
$\Rightarrow a + c = 2 b$

On squaring both sides, we get
$\Rightarrow (a + c)^{2} = 4 b^{2}$
$\Rightarrow a^{2} + c^{2} + 2 a c = 4 b^{2}$
$\Rightarrow a^{2} + c^{2} - 2 a c = 4 b^{2} - 4 a c$
$\Rightarrow (a - c)^{2} = 4 (b^{2} - a c)$

Hence, this is the answer.

Suggest Corrections

Similar questions

a, b, c

(a - c)^{2} = 4 (b^{2} - a c)

a, b, c

{(a - c)}^{2} = 4 (b^{2} - a c)

a, b

c

A . P .,

\frac{(a - c)^{2}}{b^{2} - a c} =

If a, b, c are in AP, prove that $(a - c)^{2} = 4 (b^{2} - a c)$ .

(a - c)^{2} = 4 (b^{2} - a c)

Join BYJU'S Learning Program