Question

If k+ 1,2k-1, 3k + 1 are three consecutive terms of a G.P, find the possible values of the common ratio. (JAMB)​

Answer 1

Resposta:

Olá bom dia!

The ratio (q) of a geometric progression whose elements are

X1, X2, X3,...

is equal to:

q = X2/X1

q = X3/X2

.

.

.

q = Xn / Xn-1

So with the given terms:

(2k - 1) / (k + 1) = (3k + 1) / (2k - 1)

6k² + k + 3k + 1 = 4k² - 2k - 2k + 1

6k² + 4k + 1 = 4k² - 4k + 1

6k² - 4k² + 4k + 4k + 1 - 1 = 0

-2k² + 8k = 0

2k(-k + 8) = 0

k = 0

k = -4

So:

k = 0

0 + 1 , 2(0)-1 , 3(0) + 1

{+1 , -1 , + 1, ...}

Is a g.p with ratio q = -1

k = -4

-4 + 1 , 2(-4)-1 , 3(-4) + 1

{-3 , -9 , -11, ...}

Is not a g.p.