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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.
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


Substituting the value of r in (1), we obtain

4a = a + 9

implied 3a = 9

ie, 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.


 


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