If a, b and c are in geometric progression, then $a^{2}$ , $b^{2}$ and $c^{2}$ are in _____ progression.

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

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

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

AGP

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

Open in App

Solution

The correct option is B GP
Given $a, b, c$ are in GP.
So, the common ratio between the first and second term ; second and third will be the same.
$=> \frac{b}{a} = \frac{c}{b}$

$=> b^{2} = a c$
If we square both sides, we see that
$(b^{2})^{2} = a^{2} \times c^{2}$
This means, even, $a^{2}, b^{2}, c^{2}$ are also in GP.

Suggest Corrections

Similar questions

a^{2}, b^{2}

c^{2}

\frac{- c}{2}

Let a, b, c are non-zero real numbers such that \frac{1}{a}, \frac{1}{b}, \frac{1}{c} are in arithmetic progression and a, b, -2c are in geometric progression, then which of the following statement(s) must be true?

a, b, c

d

(b - c)^{2} + (c - a)^{2} + (d - b)^{2} = (a - d)^{2}

Δ A B C

a^{2}, b^{2}, c^{2}

cot A, cot B, cot C

a^{2}, b^{2}, c^{2}

\frac{1}{(a + b)},

\frac{1}{(c + a)}, \frac{1}{(b + c)}

Join BYJU'S Learning Program