20 less than a given number K, 20 more than the given number K, and 42 more than the given number K are in continued proportion. Find the given number K. where K> 0.
Let the number be =k - 20, k + 20 and k + 42 According to the question- (k-20)/ (k + 20) = (k + 20)/ (k + 42) (k + 20)2 = (k - 20) (k + 42) (k² + 2 x k x 20 + 400) = (k² + 42k - 20k - 840) (k² + 20k + 400) = (k² + 22k-840) (22k-20k) = (840 + 400) 2k = 1240 k = 1240/2 k = 620 The value of k is 620.
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