Question

obtain the value of X for which (X+1),(X-5),(X-2) is a geometric progression.hence find the sum of the first 12 terms of the progression.​

Answer 1

If x + 1, x - 5, and x - 2 are in a geometric progression, then there is some constant r for which

x - 5 = r (x + 1)

==>   r = (x - 5) / (x + 1)

and

x - 2 = r (x - 5)

==>   r = (x - 2) / (x - 5)

Then

(x - 5) / (x + 1) = (x - 2) / (x - 5)

Solve for x :

(x - 5)² = (x - 2) (x + 1)

x ² - 10x + 25 = x ² - x - 2

-9x = -27

x = 3

It follows that the ratio between terms is

r = (3 - 5) / (3 + 1) = -2/4 = -1/2

Now, assuming x + 1 = 4 is the first term of the G.P., the n-th term a(n) is given by

a(n) = 4 (-1/2)ⁿ⁻¹

The sum of the first 12 terms - denoted here by S - is then

S = 4 (-1/2)⁰ + 4 (-1/2)¹ + 4 (-1/2)² + … + 4 (-1/2)¹¹

Solve for S :

S = 4 [(-1/2)⁰ + (-1/2)¹ + (-1/2)² + … + (-1/2)¹¹]

(-1/2) S = 4 [(-1/2)¹ + (-1/2)² + (-1/2)³ + … + (-1/2)¹²]

==>   S - (-1/2) S = 4 [(-1/2)⁰ - (-1/2)¹²]

==>   3/2 S = 4 (1 - 1/4096)

==>   S = 8/3 (1 - 1/4096)

==>   S = 1365/512