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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
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- 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
- 14 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 - 14 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 - 14 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
- 14 years agoHelpfull: Yes(1) No(3)
- none of these
- 14 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)
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