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Question
The following numbers, K + 3, K + 2, 3K – 7 and 2K – 3 are in proportion. Find K.
Solution
K + 3,K + 2,3K - 7 and 2K - 3 are in proportional then
`⇒ (K+3)/(K+2) = (3K- 7)/(2K-3)`
⇒( K +3) (2K -3) = (3k -7) (k+2)
`⇒ 2K^2 - 3K + 6K - 9= 3K^2 + 6K - 7K - 14`
`⇒ 2K^2 + 3K -9 = 3K^2 -K - 14`
`⇒ K^2 - 4K - 5 = 0`
`⇒ K^2 - 5K + K -5 = 0`
⇒ K ( K - 5 )+1( K - 5 ) = 0
⇒ K = 5 or K = -1
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