24 = 3r^3
r^3 = 24/3 = 8
r = 2
so, ...
If 3,p,q,24,are Consecutive Terms Of An Exponential Sequence,find The Values Of P And Q
5 answers
I do not know
Finish the solving
Let the common ratio be r then we have
3*r=p
P*r=q
Q*r=24
Divide the second equation by the first
P*r/3*r=q/p
P/3=q/p
P*2 =3q
3*r=p
P*r=q
Q*r=24
Divide the second equation by the first
P*r/3*r=q/p
P/3=q/p
P*2 =3q
From the third equation, we have:
q = 24/r
Substituting into the last equation yields:
P*2 = 3(24/r)
P*2 = 72/r
Multiplying both sides by 3r gives:
3Pr*2 = 216
Dividing both sides by 3P gives:
r*2 = 72/3P
r*2 = 24/P
Substituting into the second equation gives:
P*q/P = 24
q = 24
Substituting into the first equation gives:
3*r = p
In summary,
p = 6
q = 24
q = 24/r
Substituting into the last equation yields:
P*2 = 3(24/r)
P*2 = 72/r
Multiplying both sides by 3r gives:
3Pr*2 = 216
Dividing both sides by 3P gives:
r*2 = 72/3P
r*2 = 24/P
Substituting into the second equation gives:
P*q/P = 24
q = 24
Substituting into the first equation gives:
3*r = p
In summary,
p = 6
q = 24