The ratio of sum of squares of first n natural numbers to square of sum of first n natural numbers is 17:325. the value of n is (a)15(b)25(c)35(d)30 (e)none of these

• sum of 1st n natural no.s is n(n+1)/2,
  
  sum of sqaures of 1st n natural no.s is n(n+1)(2n+1)/6,
  
  The ratio of sum of squares of first n natural numbers to square of sum of first n natural numbers=n(n+1)(2n+1)*2*2/(n*n*(n+1)*(n+1))
  =2(2n+1)/(3n(n+1))
  now by hit and trial for the given options,if we put 25 then this will give 17/325
  
  therefore the answer is (b)25
  
  
• 13 years agoHelpfull: Yes(10) No(0)
• sum of square of n terms =n(n+1)(2n+1)/6 -------------- eq no.1
  square of sum of n terms =n^2(n+1)^2/4 ---------------eq no. 2
  
  eq 1/2
  
  (2n+1)2 17
  ------- = ----
  n(n+1)3 325
  
  only 25 satisfies this
• 13 years agoHelpfull: Yes(7) No(1)
• sum of squares of first natural numbers = n(n+1)(2n+1)/6
  square of sum of first natural numbers = (n(n+1)/2)^2
  thus 2(2n+1)/3n(n+1)=17/325
  comparing with the solutions for n=25 only (2n+1) becomes a multiple of 17
  ans is (b) 25
• 13 years agoHelpfull: Yes(2) No(0)
• sum of square of first n natural numbers is n(n+1)(2n+1)/2
  and sum of first n natural numbers is n(n+1)/2 so
  the ratio of above means 2(2n+1)/n(n+1)=17/325
  after solving it become 17n^2 - 1283n - 650 = 0
  the above option dont satisfy it so ans is none of these
  
• 13 years agoHelpfull: Yes(1) No(3)
• none of these
• 13 years agoHelpfull: Yes(0) No(5)
• Sum of squares of first n natural numbers is :-
  
  = n(n+1)(2n+1)/6
  
  Squares of sum of first n natural numbers
  
  = (n(n+1)/2)×(n(n+1)/2).
  
  According to the question,
  
  => n(n+1)(2n+1)/6:(n(n+1)/2)×(n(n+1)/2)=17:325
  
  Thus, by solving we get n = 25.
• 7 years agoHelpfull: Yes(0) No(0)