Find k such that k + 9, k – 6 and 4 form three consecutive terms of a G.P.

Solution:

Suppose that

a = k + 9;

b = k−6; and

c = 4;

According to the question, a, b and c are in GP, then

Making use of the property of geometric mean, we get:

b² = ac

$ {{\left( k-6 \right)}^{2}}~=~4\left( k\text{ }+\text{ }9 \right) $

$ {{k}^{2}}-12k\text{ }+\text{ }36\text{ }=\text{ }4k\text{ }+\text{ }36 $

$ {{k}^{2}}~-16k\text{ }=\text{ }0 $

$ k\text{ }=\text{ }0\text{ }or\text{ }k\text{ }=\text{ }16 $