solution is as follows: A + B + C = 18 A + C = 2B A^2 + B^2 + C^2 = 126 So in solving the first two equations we will get B= 6 then A + C = 12 OR C = (12 - A ) then substitute in the third equation we get A^2 + 6^2 + ( 12 - A )^2 = 126 we get A =3 OR A =9 The A.P will be 3 , 6 , 9 OR 9, 6 ,3
Aptitude
Topic: Progressions
in an arithmetic progression, the sum of the 1st 3 terms is 18, and the sum of the squares of these terms is 126. find the terms. how is the solution?
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