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Find the values of k for which `k+12, k-6` and 3 are in GP.

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Find the values of k for which `k+12, k-6` and 3 are in GP.
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Correct Answer - `(k=0)` or `(k=15)`
`(k-6)/(k+12)=3/(k-6) rArr (k-6)^(2)=3k+36`
`rArr k^(2) -12 k = 3k rArr k^(2) -15k =0 rArr k(k-15)=0`
`:. K=0 or k=15`.
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