Question

If p-3, p+3, 4p+3 are three consecutive terms of a geometric progression, find the possible values of p and the corresponding value of the common ratio.​

Answer 1

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.

$\dfrac{p+3}{p-3}=\dfrac{4p+3}{p+3}\\\\\\\underline{\text{Cross Multiply and solve for p:}}\\(p+3)(p+3)=(4p+3)(p-3)\\p^2+6p+9=4p^2-9p-9\\.\qquad \qquad 0=3p^2-15p-18\\.\qquad \qquad 0=3(p^2-5p-6)\\.\qquad \qquad 0=3(p-6)(p+1)\\.\qquad \qquad 0=p-6\qquad 0=p+1\\.\qquad \qquad \boxed{{p=6}}\qquad \quad\boxed{p=-1}}$

I will show that 3 is not valid below

Step 2 is to find r (common ratio).

$r=\dfrac{p+3}{p-3}\\\\\\\text{When p = 6} \rightarrow \quad \dfrac{6+3}{6-3}\quad =\dfrac{9}{3}\quad =\boxed{3}\\\\\text{When p = -1} \rightarrow \quad \dfrac{-1+3}{-1-3}\quad =\dfrac{2}{-4}\quad =\boxed{-\dfrac{1}{2}}$

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   $\checkmark$                                       -4(-1/2) = 2 $\checkmark$

9(3) = 27 $\checkmark$                                        2(-1/2) = -1 $\checkmark$

Answer 2

Answer:

Step-by-step explanation:

hello :

look this solution :