If 3,p,q,24 are consecutive terms of an exponential sequence. Find the values of p and q.

Answer 1

Let the common ratio be r. Then we have:

3 * r = p
p * r = q
q * r = 24

Dividing the second equation by the first, we get:

p * r / (3 * r) = q / p
p / 3 = q / p
p^2 = 3q

Substituting this into the third equation, we get:

(3q) * r = 24
q * r = 8

Substituting q*r=8 into the second equation, we get:

p * r = 8

Dividing this equation by the first equation, we get:

(p * r) / (3 * r) = 8 / 3
p / 3 = 8 / 3
p = 8

Substituting p=8 into p^2=3q, we get:

8^2 = 3q
64 = 3q
q = 64/3

Therefore, the values of p and q are p = 8 and q = 64/3.

Answer 2

AAAaannndd the bot gets it wrong yet again!

r^3 = 24/3 = 8
so r = 2
The sequence is 3, 6, 12, 24

Answer 3

I'm sorry for the mistake in my previous response. You are correct that the common ratio is 2. Using this, we can find the values of p and q:

3 * 2 = 6, so p = 6
6 * 2 = 12, so q = 12

Therefore, the values of p and q in the sequence are p = 6 and q = 12.