Let Sn denote the sum of first n terms of an A.P.. If S2n = 3Sn, then the ratio S3n/Sn is equal to
(a) 4 (b) 6
(c) 8 (d) 10

• Dear Admin,
  
  here is not given that it is a first natural number sum so how you can take this can you please elaborate this
  
  
  
• 10 years agoHelpfull: Yes(29) No(8)
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• 8 years agoHelpfull: Yes(9) No(2)
• b)6
  S3n=3n(3n+1)/2........(1)
  Sn=n(n+1)/2...........(2)
  Dividing (1) from (2) and substituting the value of n as 1, we get the ans as 6.
• 12 years agoHelpfull: Yes(8) No(8)
• solve using Sn=n/2[2a+(n-1)d]
  value of d=2a/(n+1) put it in required equation.
  ans 6.
• 10 years agoHelpfull: Yes(8) No(3)
• (b) 6
  
  Sn denote the sum of first n terms of an A.P.. If S2n = 3Sn, then the ratio
  S3n / Sn is equal to 6.
  and d = a/(1+n)
• 12 years agoHelpfull: Yes(6) No(7)
• dear ravindra
  if the result is true for any general case then it is true for 1st natural numbers as well;
  you can try it using any AP
• 10 years agoHelpfull: Yes(4) No(4)
• ans will be 6.
  
• 9 years agoHelpfull: Yes(3) No(2)
• let n=1
  then sn=1
  s2n=1+2=3
  s3n=1+2+3=6
  s3n/sn=6/1=6
• 9 years agoHelpfull: Yes(3) No(2)
• solve using Sn=n/2[2a+(n-1)d]
  value of d= -2a/(n+1) put it in required equation.
  ans 6.
• 10 years agoHelpfull: Yes(2) No(2)
• taking A.P 1,2,3
  S1=1, S2 = 3 , S3=6
  S2= 3*S1
  S3/S1 = 6/1=6
• 10 years agoHelpfull: Yes(2) No(3)
• Hi saurabh, i had taken 2 to 3 A.P Series but it does not satisfy the rule......
  
• 10 years agoHelpfull: Yes(2) No(3)
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• 8 years agoHelpfull: Yes(2) No(1)
• @Ravindra admin approach is right as the AP can be sum of 1st n terms or we can assume as sum of 1st natural numbers ...but you can go with another approach like this...
  A/q s2n=3sn
  =>2n/2(2*a+(2n-1)d)=3*n/2(2a+(n-1)d)
  =>a=d*(n+1)/2
  now,
  s3n/sn=
  3n/2(2*a+(3n-1)d)]/[n/2(2*a+(n-1)d)]
  =>3*2[a+(3n-1)d/2] / 2*[a+(n-1)d/2]
  =>3*[a+(3n-1)(a/n+1)] / [a+(n-1)(a/n+1)]
  =>3*[1+(3n-1)/(n+1)] / [1+(n-1)/(n+1)]
  =>3*4n/2n
  =>3*2=6(ans)
• 6 years agoHelpfull: Yes(1) No(0)
• S3n=S2n+Sn
  S3n=3Sn+Sn
  S3n=4Sn
  S3n/Sn=4
  
  why this approach is wrong??
• 8 years agoHelpfull: Yes(0) No(1)
• Because s3n != s2n + sn
• 8 years agoHelpfull: Yes(0) No(1)
• if S2n=3Sn
  then why we can not deduce S3n=4Sn?
• 7 years agoHelpfull: Yes(0) No(1)
• sn=n(n+1)/2
  2n(2n+1)/2=3n(n+1)/2
  n=1
  (3n(3n+1)/2)/n(n+1)/2=6
• 6 years agoHelpfull: Yes(0) No(1)
• Sn = n*(n+1)/2
  given: S2n = 3Sn
  
  2n(2n+1)/2 = 3n(n+1)/2
  2(2n+1)=3(n+1)
  n=1
  
  S3n/Sn = (3n(3n+1)/2)/(n(n+1)/2)=(3*4/2)/(1) = 6
• 3 years agoHelpfull: Yes(0) No(0)