# 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.

Given: a3 = a1 + 9 and a2 = a4 + 18

Let a1 = a, a2 = ar, a3 = ar2, a4 = ar3

Here,

a3 = a1 + 9

ar2 = a + 9

ar2 – a = 9

a(r2 – 1) = 9 – 1

and,

a2 = a4 + 18

ar= ar3 + 18

ar – ar3 = 18

ar(1 – r2) = 18 – 2

Divide eq –2 by eq –1

r = – 2

Substitute r in eq—1

a × ((-2)2 – 1) = 9

a × (4 – 1) = 9

a × (3) = 9

a = = 3

G.P is : 3 , 3(– 2), 3(– 2)2, 3(– 2)3

3, —6 , 12 , —24

G.P is 3, —6, 12, —24

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