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 4^th 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.