Question
The equation x2 β px β 60 = 0, has two
roots βaβ and βbβ such that (a β b) = 17 and p > 0. If a series starts with βpβ such that the consecutive terms are 4 more than the preceding term is formed, then find the product of 2nd and 4th terms of such series.Solution
Given, x2 + px β 60 = 0 Since, sum of roots = -(-p)/1 So, a + b = p Since, product of roots of the equation = -(60/1) Therefore, ab = -60β¦β¦. (1) Or, b = (-60/a) And, a β b = 17β¦β¦. (2) Putting the value of βbβ in equation (2), we get a2 β 17a + 60 = 0 => a2 β 12a β 5a + 60 = 0 => a(a β 12) β 5(a β 12) = 0 => (a β 12)(a β 5) = 0 => a = 12, 5 When a = 12, then b = -5(-60/12) And, when a = 5, then b = -12(-72/5) Therefore, a + b = 12 + (-5) = 7 = p Therefore, series will be 7, 11, 15, 19 required product = 11 Γ 19 = 209
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