Two series are 16, 21, 26.... And 17, 21, 25.....What is the sum of first hundred common numbers
(a) 101100
(b) 110100
(c) 101110
(d) 110101

• Ans - a)101100
  Expln:
  By Expanding both series like
  16 21 26 31 36 41 46 51 56 61.......
  17 21 25 29 33 37 41 45 49 53 57 61...
  we can see that common terms are:21 41 61 81 101....
  it is in the form of AP wit d=20 we r askd to find sum of first 100.
  so the solution is 100/2(2*21+(100-1)20)= 101100
• 13 years agoHelpfull: Yes(17) No(2)
• (a)101100 ans.
  explanation:-
  The series of first hundred common numbers will be:
  21,41,61,81,...,100th
  since,tn=a+(n-1)*d
  where, a=1st term=21
  d=common difference=41-21=61-41=...=20
  so,
  tn=21+(100-1)*20=2001
  therefore,
  last term(l)=2001
  hence,
  Sn=n/2(a+l)
  S100=100/2(21+2001)=50*2022=101100 ans.
• 12 years agoHelpfull: Yes(10) No(0)
• Ans - a)101100
  Both series are Arithmatical series with common difference 4 and 5 respectively. So the terms of the series are :
  16 21 26 31 36 41 46 51 56 61.......
  17 21 25 29 33 37 41 45 49 53 57 61...
  The series with common numbers of both the series is an Arithmatical series with first term 21 and common difference 20.
  So the series of common numbers is 21 41 61 81 101....
  The sum of first 100 numbers = 100/2(2*21+(100-1)20)= 100*(21+990)
  = 100*1011 = 101100
• 13 years agoHelpfull: Yes(7) No(1)
• 16,21,26....here diff b/w two term is 5
  17,21,25....here diff b/w two term is 4
  5*4=20=4*5
  so,21,41,61,81,101......
  it is in A.P.
  so sum is=100/2(2*21+(100-1)*20)=50(42+1980)=101100
• 13 years agoHelpfull: Yes(6) No(1)
• =50(42+1980)
  =101100
• 13 years agoHelpfull: Yes(1) No(7)
• 1st series is with increment of 5 and 2nd with 4 and lcm of 4 &5 is 20 so it repeats after 20 increment with previous and find last no. with n=a+(n-1)d and calculate sum with sum=n(1st term+last term )/2
• 10 years agoHelpfull: Yes(0) No(0)
• for what reason we find the last number??? pls help me frnds
• 10 years agoHelpfull: Yes(0) No(0)
• can not be determined...bcoz any 1 among C and F can be third to left from D
  
• 10 years agoHelpfull: Yes(0) No(0)
• 16,21,26....here diff b/w two term is 5
  17,21,25....here diff b/w two term is 4
  5*4=20=4*5
  so,21,41,61,81,101......
  it is in A.P.
  so sum is=100/2(2*21+(100-1)*20)=50(42+1980)=101100
  
• 10 years agoHelpfull: Yes(0) No(0)
• number will be 21,41,61,81,101.......
  a=21
  n=100
  d=20
  so,sum=n/2(2a+(n-1)d)=
  100/2(2*21+99*20)=101100...
• 10 years agoHelpfull: Yes(0) No(0)
• common term
  16 21 26 31 36 41 46 51 56 61.......
  17 21 25 29 33 37 41 45 49 53 57 61... so n=100,a=21,d=20..formula -----sum=n/2(2*a +(n-1)*d)
  sum=100/2(2*21+(100-1)*20)=101100 ans
  
• 9 years agoHelpfull: Yes(0) No(0)
• 16, 21, 26, & 17, 21, 25,
  Common number from the two series
  16 + 5n = 17 + 4m → 5n -1 = 4 m
  The first number is 21
  And the second :at n = 5 and m = 6
  The number = 16+25= 41
  
  The series become 21, 41, 61 U₁₀₀= 21+99 . 20= 2001 S₁₀₀= ½.100(21+2001)= 50 .2022 = 101100
• 9 years agoHelpfull: Yes(0) No(0)