Two series are 16,21,26.... and 17,21,25.....What is the sum of first hundred common numbers

• Sum of first hundred Common Number will be 101100.
  
  1st Series increasing by 5
  2nd Series increasing by 4
  
  By extending two series, we get,
  
  1st Series = 16 21 26 31 36 41 46 51 56 61 66 71 76 and so on..
  2nd Series = 17 21 25 29 33 37 41 45 49 53 57 61 63 and so on..
  
  By above series, we have the common terms
  
  21 41 61 81 101 121 and so on.. (Form of A.P. where d=20)
  
  To finding Sum of first 100 common terms, we use the below formula
  [n/2(2a+(n-1)d]
  
  n=100 ; a=21; d=20; (n=number of terms; a=Starting common term; d=difference)
  
  Substituting those values in the formula,
  
  [100/2*(2*21+(100-1)*20)]
  [50*(42+(99)*20)]
  [50*(42+1980)]
  [50*2022]
  101100.
• 12 years agoHelpfull: Yes(22) No(0)
• The pattern of the series with the given condition would be
  21,41,61, up to 100 terms
  This is an AP with a common difference 20
  The 100th term is 2001( use the formula, a+(n-1)d
  Now, For summation of AP series, Use the formula, n/2(2a+(n-1)d
  WE would get the answer as 101100
• 12 years agoHelpfull: Yes(4) No(0)
• common no. in series are 21,41,61.......
  hence common difference=20
  no of terms=100
  first term=21
  hence sum=n/2[2a+(n-1)d]
  =101100
• 12 years agoHelpfull: Yes(2) No(0)
• Series one - 16, 21, 26, 31, 36, 41,...
  Series two - 17, 21, 25, 29, 33, 37, 41,...
  Common terms 21, 41,...
  this is also a series in AP
  Sum of n terms = n/2{2*a+ (n-1)*d}
  where a = first term
  d = difference between two consecutive terms (2nd term - 1st term)
  Using the formula-
  sum of 100 terms = 100/2{2*21 + (100 - 1)*10}
  = 101100.
• 12 years agoHelpfull: Yes(1) No(0)
• ans:101100
  
  the series formed by common numbers is 21,41,61...
  we use the formula n/2(2a+(n-1)d) to solve this.
  100/2(2*21(100-1)20) = 101100
• 12 years agoHelpfull: Yes(1) No(1)
• common difference of series are 4,5.
  common term exists for every 4*5=20 numbers..........
  21,41,61,81,........which are in a.p.
  sum of 100 numbers=n/2(2a+(n-1)d)
  =100/2(2*21+(100-1)20)=101200
• 11 years agoHelpfull: Yes(1) No(1)