The last digit of the expression 4+92+43+94+45+96+....+499+9100 is.

• 3 :
  43+45+.........499 are in a.p. find sum of digits using n/2(2a+(n-1)d)
  last digit of this sum is 9. similarly 92+94+...........9100.,last digit of this sum is 0.. soo 9+4+0 = 13. 3 is last digit.
• 10 years agoHelpfull: Yes(29) No(4)
• 7 :
  43+45+.........499 are in a.p. find sum of digits using n/2(2a+(n-1)d)
  last digit of this sum is 9. similarly 92+94+...........9100.,last digit of this sum is 4.. soo 9+4+4 = 17. 7 is last digit.
• 10 years agoHelpfull: Yes(10) No(8)
• Yeah Keerthana! 3 is the right answer!
• 10 years agoHelpfull: Yes(6) No(3)
• In the sum of 92+94+...9100 , the last digit of this sum is "0" ryt??
• 10 years agoHelpfull: Yes(5) No(6)
• Two different AP series exist here :
  
  1) 43+45+47+....+499
  2)92+94+96+....+9100
  
  499=43+(n-1)2
  =>n=229
  Sum=229*(43+499)/2=Last digit will be 9
  
  9100=92+(n-1)2
  =>n=4459
  Sum=4459*(92+9100)/2=Last digit will be 4
  
  So sum of last digits is 4+9+4=17
  Hence last digit is 7
  
  
  
• 9 years agoHelpfull: Yes(2) No(2)
• 43+45+...+499 are in AP ...after finding out the no of terms using tn=a+(n-1)d, we get the sum of the AP series....the last digit of which comes to 9...by using the same process for 92+94+....+9100, the last digit for it comes to 0..thus the last digit of the whole series becomes 9+0=9. Thus answer is 9.
• 9 years agoHelpfull: Yes(1) No(3)
• 92+94+96+...then 43+45+47+....the series end with a even number. 501
  
• 7 years agoHelpfull: Yes(0) No(0)
• 43+45+………499 are in a.p.
  find sum of digit using n/2(2a+(n-1)d)
  last digit of this sum is 9.
  Similarly 92+94+………..9100.
  Last digit of this sum is 0.
  So 9+4+0 = 13.
  3 is last digit.
  
• 7 years agoHelpfull: Yes(0) No(0)