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 term by 18.

Answer 1

Let ‘a’ be the first term and ‘r’ be the common ratio of a G.P.

Four numbers in G.P.be a, ar, ar²  and ar³ 

Given that T₃ = T_x + 9 and T₂ = T₄ + 18. 

i.e., ar² - a + 9 …………….. (1)  and 

ar = ar³ +18 ……………… (2) 

From (1), we get 

ar²  – a = 9 

⇒ a(r² - l) = 9   …(3) 

From (2), we get 

ar – ar³  = 18 

⇒ nr(1- r²) = 18 …(4) 

Dividing (4) by (3), we get 

r = 2 ⇒ r = -2 

Substituting r = – 2 in (1), we get 

a( 4) = a + 9 

⇒ 3a = 9 

⇒ a = 3 Hence, four numbers are, 

⇒ 3, 3(-2), 3(-2)² , 3(-2)³  i.e., 3,-6,12,-24