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

Explanation:

hello :

look this solution :

Answer 2

Answer: p = 6, r = 3 or p = -1, r = -1/2

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.

`(p+3)/(p-3)=(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=(p+3)/(p-3)\\\\\\\text{When p = 6} \rightarrow \quad (6+3)/(6-3)\quad =(9)/(3)\quad =\boxed{3}\\\\\text{When p = -1} \rightarrow \quad (-1+3)/(-1-3)\quad =(2)/(-4)\quad =\boxed{-(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`