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Question
Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
Solution
Let the geometric series be a, ar, ar2, ar3,...
Third term = ar2, first term = a
∴ ar2 – a = 9 …........(i)
Second term = ar, fourth term = ar3
ar – ar3 = 18 ….........(ii)
Dividing equation (i) by (ii), we get
`("a"("r"^2 - 1))/("a"("r" - "r"^3))`
= `9/18`
= `1/2`
or 2(r2 − 1) = r − r3
∴ r3 + 2r2 − r − 2 = 0
or (r − 1) (r + 1) (r + 2) = 0
or r = 1, −1, −2 if r = −2,
From equation (i), a(4 − 1) = 9
∴ a = 3
∴ 4th terms of the geometric progression 3, −6, 12, −24.
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