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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.
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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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