Question 97
Let T be the set of integers {3,11,19,27,…451,459,467} and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in S is
Solution
No. of terms in series T , 3+(n-1)*8 = 467 i.e. n=59.
Now S will have atleast have of 59 terms i.e 29 .
Also the sum of 29th term and 30th term is less than 470.
Hence, maximum possible elements in S is 30.
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