sum of 44,42,40,...?

• The series is as 44,42,40, ....(Its a infinite Series)
  The max value can be obtained when the series will have +ve elements in it.
  So the max sum will be (44+ 42+ ....+ 2) which is an AP with common Difference -2.
  So the sum = n(1st + Last term)/2. Here can be obtain by the formula Tn = a + (n-1)*d. Which will give n = 22. By putting Tn = 2,a = 44 & d = -2
  So Sum = 22(44+2)/2 = 506.
• 11 years agoHelpfull: Yes(12) No(0)
• SUM IS MAX WHEN LAST TERM = 0
  =>COMMON DIFFERENCE=-2
  SUM=N/2(FIRST+LAST)
  23/2(44+0)
  =>506
  
• 11 years agoHelpfull: Yes(9) No(0)
• The sum is maximum if the series does not goes to (-ve)
  hence the given series is 44,,42,40........0
  Let no. of terms in this series is n
  0=44+(n-1)(-2)
  hence n=23
  now the sum of these 23 terms of series is
  23/2[2*44+22*(-2)]
  23/2*44
  23*22
  =506
  
• 11 years agoHelpfull: Yes(8) No(0)
• 44,42,40,…
  This is a decreasing AP. If you leave it like that, it will go in to negative terms -2,-4,….
  But to make the sum maximum, we need to stop at ‘0’. Otherwise the –ve terms make the sum decrease.
  So the series is 44,42,40,…..0.
  44+42+40+…+0 = 2(22+21+20+…0) = 2(sum of first 22 integers) = 2(n*(n+1)/2), where n =22
  => Sum = 2*(22*23/2) = 22*23 = 506
  Ans:506
• 10 years agoHelpfull: Yes(1) No(0)