A) 10
B) 11
C) 12
D) None of these
Correct Answer: B
Solution :
Let three number of A.P. \[a-d,\,a,\,\]\[a+d\] Sum = 18, and \[{{(a-d)}^{2}}+{{a}^{2}}+{{(a+d)}^{2}}=58\] \[a-d+a+a+d=18\] \[a=6\] and \[{{(6-d)}^{2}}+36+{{(6+d)}^{2}}=158\] = \[36+{{d}^{2}}+36+{{d}^{2}}=122\]\[=2{{d}^{2}}+72=122\] \[=2{{d}^{2}}=50\] Þ \[d=5\]. Hence Numbers are 1, 6, 11, i.e. maximum number is 11.
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