The sum of how may terms of the series 6 + 12 + 18 + 24 + ... is 1800 ?

• This is an A.P. in which a = 6, d = 6 and Sn = 1800
  
  Then, n/2[2a + (n - 1)d] = 1800
  
  n/2[2 x 6 + (n - 1) x 6] = 1800
  
  => 3n (n + 1) = 1800
  
  => n(n + 1) = 600
  
  => n2 + n - 600 = 0
  
  => n2 + 25n - 24n - 600 = 0
  
  => n(n + 25) - 24(n + 25) = 0
  
  => (n + 25)(n - 24) = 0
  
  => n = 24
  
  Number of terms = 24.
• 12 years agoHelpfull: Yes(13) No(2)
• SOL:24
  
  6+12+18+24+...............=1800
  =>6(1+2+3+4+.............)=1800
  =>(1+2+3+4+.............)=300
  =>N(N+1)/2=300 SOLVING THIS N=24
• 12 years agoHelpfull: Yes(5) No(0)
• 24 terms of the given series gives sum 1800
  since given series is in A.P with common difference (d) = 6 .
  let no. of terms = n .
  then
  sum = n/2 [2*(first term) + (n - 1)*d]
  on solving we get n = 24.
• 12 years agoHelpfull: Yes(1) No(0)