Step-by-step explanation:
since, a,b,c are in GP.
[tex] \mathsf{\frac{b}{a} = \frac{c}{b}} [/tex]
upon cross multiplying, we get
b² = ac . . . . . (¡)
We have to prove if a^n, b^n and c^n are also in GP.
We'll simply write these numbers in the bove format and if both the sides are equal in the end, it's proved else it's not.
[tex] \mathsf{ \frac{ {b}^{n} }{ {a}^{n} } = \frac{ {c}^{n} }{ {b}^{n} } }[/tex]
once again cross multiplying
[tex] \mathsf{ \ {(b {}^{2}) }^{n} = {a}^{n}{c}^{n} }[/tex]
[tex] \mathsf{ \ {(b {}^{2}) }^{n} = {(ac)}^{n} }[/tex]
from eqn. (¡) b² = ac
[tex] \implies \mathsf{ {(ac)}^{n} = {(ac)}^{n} }[/tex]
both the sides are equal, sooo our assumption was correct a^n, b^n, c^n are in GP.
