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16. Let a,b,c be positive integers such that is an integer. If a.b.c are in geometric progression and thea2+a-14arithmetic mean of a,b,c is b + 2, then the value of isISa+1JEE (Advanced) 2014, Paper-1, (3, oy60]17. Sunnose that oll tho to |
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Answer» Let b=na. Since b/a is an integer and both b and a are positive, n is also a positive integer. Since the numbers are in geometric progression, it follows that:c=n^2 a Given that their arithmetic mean is b+2 (=na+2), it means:(a + na + n^2 a)/3 = na+2 Simplifying,n^2 -2n + (1 - 6/a) = 0 Finding the roots of this quadratic equation in n,n =1 ± 0.5√(24/a) [using quadratic formula]Since n is an integer, 24/a has to be a perfect square. Hence the only possible value of a is 6, which gives n=2 So the values of a, b, c are 6, 12, 24, respectively.Now putting the values we get,a2+ a -14/a+1=6^2+6-14/7=4 |
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