Question

Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

Answer 1

Let a be the first term and r be the common ratio of the G.P.
a1 = a, a2 = ar, a3 = ar2 , a4 = ar3
By the given condition,
a3 = a1 + 9
⇒ ar2 = a + 9 … (1)
a2 = a4 + 18
⇒ ar = ar3 + 18 … (2)
From (1) and (2), we obtain
a(r2-1) = 9 … (3)
ar (1-r2 ) = 18 … (4)
Dividing (4) by (3), we obtain
$\frac{ar (1 - r^2) }{a (r^2 - 1) }= \frac{18 }{9}$
⇒ - r = 2
⇒ r = - 2
Substituting the value of r in (1), we obtain
4a = a + 9
⇒ 3a = 9
∴ a = 3
Thus, the first four numbers of the G.P. are 3, 3(-2), 3(-2)2 , and 3 (-2)3 i.e., 3 ¸-6, 12, and -24.