Answer: p = 6, r = 3 or p = -1, r = -1/2
Step-by-step explanation:
Since p - 3, p + 3, and 4p + 3 are consecutive terms in a geometric progression, then the proportion of the 2nd term over the 1st equals the 3rd term over the 2nd term.
Step 1 is to find p.
[tex]\dfrac{p+3}{p-3}=\dfrac{4p+3}{p+3}\\\\\\\underline{\text{Cross Multiply and solve for p:}}\\(p+3)(p+3)=(4p+3)(p-3)\\p^2+6p+9=4p^2-9p-9\\.\qquad \qquad 0=3p^2-15p-18\\.\qquad \qquad 0=3(p^2-5p-6)\\.\qquad \qquad 0=3(p-6)(p+1)\\.\qquad \qquad 0=p-6\qquad 0=p+1\\.\qquad \qquad \boxed{{p=6}}\qquad \quad\boxed{p=-1}}[/tex]
I will show that 3 is not valid below
Step 2 is to find r (common ratio).
[tex]r=\dfrac{p+3}{p-3}\\\\\\\text{When p = 6} \rightarrow \quad \dfrac{6+3}{6-3}\quad =\dfrac{9}{3}\quad =\boxed{3}\\\\\text{When p = -1} \rightarrow \quad \dfrac{-1+3}{-1-3}\quad =\dfrac{2}{-4}\quad =\boxed{-\dfrac{1}{2}}[/tex]
Check:
p = 6 p = -1
p - 3: 6 - 3 = 3 -1 - 3 = -4
p + 3: 6 + 3 = 9 -1 + 3 = 2
4p + 3: 4(6) + 3 = 27 4(-1) + 3 = -1
3(3) = 9 [tex]\checkmark[/tex] -4(-1/2) = 2 [tex]\checkmark[/tex]
9(3) = 27 [tex]\checkmark[/tex] 2(-1/2) = -1 [tex]\checkmark[/tex]
