sum of 100 terms common to progressions 17,21,25,29..... and 116,21,26,......

• first term 21;
  common diff=20;
  sum=100/2(2*21+99*20)=101100
• 10 years agoHelpfull: Yes(15) No(0)
• Terms common to both the expressions are:
  
  21,41,61,81.....
  
  Here First term(a)=21, Common Difference(d)=20, n=100
  
  So sum of terms in the AP = (n/2)*(a+(n-1)*d)
  = (100/2)*(21+(100-1)*20)
  = 50*(21+99*20)
  = 50*2001
  = 100050
  
• 10 years agoHelpfull: Yes(2) No(10)
• ans is....4220
  
  common term is 21,41,61......401
  so total term is 20 and d=20
  so sum is 20/2(21+401)=4220
• 10 years agoHelpfull: Yes(2) No(5)
• average of first and last term = (17 + (17 +(100*4)/2 = (17+417)/2 = 217
  sum of 100 terms = 217 * 100 = 21700
• 10 years agoHelpfull: Yes(0) No(2)
• 17 21 25....
  21 26 31...
  common terms of both series r 21 41 61 so in AP wit cd 20
  100th term is 2022
  therefore sum is 100/2(21+2022)=10100
• 10 years agoHelpfull: Yes(0) No(5)
• first term is 21 then 41,61 upto 100 term the 100 term is 2001.
  therefore sum is..
  100/2{21+2001}
  
  2022*50= 101100 ans.
• 10 years agoHelpfull: Yes(0) No(0)