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 ??

1) 32
2) 28
3) 29
4) 30

• from the given we see that (first term+last term)=(467+3)=470. Now according to property of AP or by just seeing the series we can determine that (second+second last)=(third+third last) and so on all will be 470 upto the middle term. So finding the no of terms in the series it would be
  
  a=3 d=8
  
  No of terms={(last term-first term)/Common difference}+1 ie. {(467-3)/8}+1 = 59
  
  so from what is mentioned above we have ( 29 terms , middle term, 29 terms)
  
  so the first 29 and last 29 terms form 470 so we exclude the last 29 terms the middle term will not be 470
  so maximum possible elements in subset s = 29+1=30 Elements
• 8 years agoHelpfull: Yes(0) No(0)
• Set t is an AP with a=3 and d=8, sum of any two elements must not exceed 470 then find the order of term which is half of 470, i.e. am = a + (m-1)d = 235, you will get m = 30, now sum of any two elements from order 1 to 30 will be less than 470 and hence answer is 30.
• 4 years agoHelpfull: Yes(0) No(0)