Q. Sum of all triple digit numbers divisible by 7.

• The first three digit no. divisible by 7 is 105 & last is 994
  so numbers are 105,112,119,...,994
  sum=105+112+119+...+994
  series is AP with first a=105,d=7,l=994
  994=105+(n-1)*7
  or 889=(n-1)*7
  or 127=n-1
  or n=128
  so sum=(n/2)*(a+l)
  =(128/2)*(105+994)
  =70336
• 11 years agoHelpfull: Yes(10) No(0)
• The first one is 105, which can be written as 7*15
  The last one is 994, which can be written as 7*142
  
  So, a total of 128 numbers
  The average is (105 + 994)/2 = 1099/2
  So, total is 128 *1099/2 = 64*1099 = 74336
  
  Answer is 74336
• 11 years agoHelpfull: Yes(0) No(0)
• The first one is 105, which can be written as 7*15
  The last one is 994, which can be written as 7*142
  So from 105 to 994, 128 numbers are present.
  
  Ans is [(1st number+Last numbber)*Total number]/2
  =[(105+994)*128]/2=74336
• 11 years agoHelpfull: Yes(0) No(0)
• all three digits no. divisible by 7 are 105,112,119......994
  the first no. 105=7*15
  the last no. 994=7*142
  
  the total no. of three digit numbers divisible by 7 are ((142-15)
  +1)=128....(1 is added because both 12 and 142 are also included)
  
  now putting the formula for AP
  sum=n/2(a+l)
  =128/2(105+994)
  =70336
  
  
  
  
  
• 11 years agoHelpfull: Yes(0) No(0)
• 105,114.....998
  common diff=7
  
  994=a+(n-1)d
  994=105+(n-1)7
  889+7 = 7n
  n=128
  
  sum of these terms=(n/2)*(a+l)
  =64*110=70336
• 11 years agoHelpfull: Yes(0) No(0)