If 2 ,   7 ,   9 and 5 are subtracted respectively from four numbers in geometric progression, then the resulting numbers are in arithmetic progression. The smallest of the four numbers is

Step-by-step solution

Let the numbers be a , ar , ar 2 , ar 3 , ⇒ a - 2 , a r - 7 , a r 2 - 9 and a r 3 - 5 are in A . P . ⇒ a - 2 + a r 2 - 9 = 2 a r - 7 and a r - 7 + a r 3 - 5 = 2 a r 2 - 9 ⇒ a + a r 2 = 2 a r - 3 and a r + a r 3 = 2 a r 2 - 6 ⇒ 1 + r 2 = 2 r - 3 a and 1 + r 2 = 2 r - 6 a r ⇒ 3 a = 6 a r ⇒ r = 2 ⇒ 3 a = 2 r - 1 - r 2 = 4 - 1 - 4 = - 1 ⇒ a = - 3 So, the numbers are – 3 , – 6 , – 12 , – 24