find X if the numbers X+3, 5x-3, and 7x+3 are three consecutive terms of a G.P of positive terms. with this value X and given the X+3, 5x -3 and 7x +3 are the third, fourth and fifth terms of the G.P.. find the sum of the first 8 terms of the progression.

Answer 1

If GP, then

(5x-3)/(x+3) = (7x+3)/(5x-3)
25x^2 - 30x + 9 = 7x^2 + 24x + 9
18x^2 - 54x = 0
6x(3x - 9) = 0
x= 0 or x = 3

if x = 0
3rd term = 3
4th term = -3
5th term = 3
..
1st term = 3
2nd term = -3
sum of any even number of terms = 0

if x = 3,
3rd term = 6
4th term = 12
5th term = 24
so r = 2
t(3) = ar^2 = 6
a = 3/2

sum(8) = a(r^8 - 1)/(r-1)
= (3/2)((2^8 - 1)/(3/2 - 1)
= (3/2)(255)/(1/2)
= 765

Answer 2

The first answer is ok, but for the second answer I got 384 and the book says it's 382.5. I'm really confused

Answer 3

Well, you got mistakes in the second part