if the sum of 2000 consecutive integer is 5000 then what will be the sum of each digit for last term.
option:
A-4
B-3
C-5
D-none of these

• Sn=n/2{2a+(n-1)d}
  
  Sum is 5000 of 2000 consecutive integers.
  so, d= 1 (since nos are consecutive), n=2000, Sn =5000
  5000=2000/2{2a+(2000-1)*1}
  solving, a= -997
  NOW,
  apply Sn=n/2(a+l)
  5000=2000/2(-997+l)
  l=1002 (last term)
  so summation is 1+0+0+2= 3
  --option B- 3
• 9 years agoHelpfull: Yes(66) No(0)
• I AM PROUD OF @KUMAR AVINASH
• 9 years agoHelpfull: Yes(5) No(2)
• 2502 last term. ans 9 i.e D
• 9 years agoHelpfull: Yes(1) No(8)
• jiyo chacha tum to hila diye!!
• 9 years agoHelpfull: Yes(1) No(1)
• 3 last term will be 1002
• 9 years agoHelpfull: Yes(0) No(5)
• n=2000; sn=5000
  n/2(2*a+(n-1)*d)=5000
  solving a=-997
  t2000=1002
  ans=3
  
• 9 years agoHelpfull: Yes(0) No(0)
• 1000(2*a+1999)=5000 (d=1)
  from above a=-997 which is 1st term
  now, -997+1999=an
  an=1002
  sum of digit of 1002 is 3
• 8 years agoHelpfull: Yes(0) No(0)
• chatur right.........
• 7 years agoHelpfull: Yes(0) No(1)