Question

If a b c are in ap then prove that k^a, k^b, k^c are in gp

Answer 1

Given---> a , b , c are in AP.

To prove ---> kᵃ , kᵇ , kᶜ are in GP .

Proof ---> If a , b , c are in AP then

b - a = c - b

=> b + b = a + c

=> 2 b = a + c

Now if p , q and r are in GP then

=> q / p = r / q

=> q × q = r × p

=> q² = p r

it is the condition of three terms are in GP

Now we have to prove that kᵃ , kᵇ , kᶜ are in GP for this it is sufficient to prove that

kᵃ × kᶜ = ( kᵇ )²

Now we proceed with

kᵃ × kᶜ = kᵃ⁺ᶜ

Putting a + c = 2b in exponent

= k²ᵇ

kᵃ × kᶜ = ( kᵇ )²

Clearly kᵃ , kᵇ , kᶜ are in GP .

Answer 2

Answer:

Please see the attachment