The sum of first 40 terms 1,6, 7,12, 13, 20... ?

• in this series two series ,
  1st is 1,7,13...115
  this series sum=n/2[2a+(n-1)d]=1160.
  and second series is
  6,12,20,30,42,54,....432
  sum=3390.
  so,total sum=3390+1160=4550
• 10 years agoHelpfull: Yes(7) No(14)
• The given series consists of 2 APs...
  AP1: 1,7,13... use sum formula => sum = n/2[2a+(n-1)d] , here a=1 n=20 d=1;
  so sum = 1160.
  
  AP2: 6,12,20...
  or 2*3, 3*4, 4*5.....
  ****IMPORTANT****
  this is AP is further broken down into 2 APs
  
  AP1 : 2,3,4,5..
  AP2 : 3,4,5,6..
  
  now to find the sum..we will use summation (sigma)....
  
  sigma(n=1 to 20) [a1+(n-1)d]*[a2+(n-1)d] ,,,,,,,here a1=2 and a2=3
  
  so it becomes summation of sigma (n=1 to 20) n^2 + 3n + 2...
  
  this can be solved using summation n^2 = n(n+1)(2n+1)/6
  summation 3n = 3n(n+1)/2
  summation 2 = 2n
  
  
  put n=20 in all...
  sum = 3540..
  
  Ans. is 1160 + 3540 = 4700
  
  Hope it helps..!!
  
• 9 years agoHelpfull: Yes(4) No(1)
• sorry my above approch is wrong...
• 10 years agoHelpfull: Yes(3) No(0)
• Answer is 4700.. Verified!
• 10 years agoHelpfull: Yes(3) No(1)
• 6,12,20....it is nt an AP.so hw to get the sum of 20 terms?plz explain
• 10 years agoHelpfull: Yes(2) No(0)
• the 2 series are:-
  1,7,13,..
  series is in ap so sum is = n/2[2a+(n-1)d]=1160 (n=20 and a=1)
  the second series is:-
  6,12,20,30,...
  the difference of the above series is in ap ie the difference is :- 6,8,10,..
  so the last difference of terms will be= a+(n-1)d=42 (a=6 and n=19 and d=2)
  so the second series is:- 6,12,20,30,....460
  sum of the second series=3510
  total sum=3510+1160=4670
  
  
• 10 years agoHelpfull: Yes(2) No(3)
• sum of first series=1160
  for second 6,12,20....462=3540
• 10 years agoHelpfull: Yes(2) No(0)
• 19...7-1=6,13-7=6,19-13=6
• 10 years agoHelpfull: Yes(0) No(7)
• 1095814995
• 10 years agoHelpfull: Yes(0) No(3)
• dividing the above sereis into two groups
  1,7,13,.....139(total 20 terms)
  6,12,18,....126(total 20 terms)
  using formula sum of terms =n/2(2a+(n-1)d)
  for first sereis n=139,a=1,d=6 substituting in above formula we get 57685
  again for second sereis n=126,a=6,d=6 sum=48006
  sum of two sereis=57685+48006=105691 is the answer
• 10 years agoHelpfull: Yes(0) No(11)
• 7/10
  in above i explain but mistake in total term
  
• 10 years agoHelpfull: Yes(0) No(0)
• we have two series of AP
  first--- 1 7 13......
  2nd-- 6 12 20......
  calculate individual sum of each
  first-- Sum=560
  2nd--- 660
  Total = 560+660=1120
  ans is 1120
• 10 years agoHelpfull: Yes(0) No(2)
• m not getting it..pls explain?
• 10 years agoHelpfull: Yes(0) No(0)
• 4700
  sum of ap + sum of 6,12,20
  
• 10 years agoHelpfull: Yes(0) No(0)
• in the given question there are two different series:
  1,7,13....
  last term/20th term if the series: 1+(20-1)x6=115
  so the series is 1,7,13....115
  sum of the series is 20/2(1+115)=1160
  
  now consider the second series:6,12,20....
  we could break it in to two series:
  1x6,2x6+0,3x6+2,4x6+4,5x6+6.....
  which could be break down in to
  6,12,18....120(up to 20th term)
  Sum:1260
  and
  2,4,6....36(up to 18th term)
  sum:162
  TOTAL SUM:2582(adding all 3 sum)
• 10 years agoHelpfull: Yes(0) No(2)
• ans=4700..its correct verifiedand
• 10 years agoHelpfull: Yes(0) No(2)
• 4700
  sum of ap + sum of 6,12,20
  
• 9 years agoHelpfull: Yes(0) No(1)
• how dis you find the sum of second series?? i.e 6,12,20,,.....?? please elaborate..
• 9 years agoHelpfull: Yes(0) No(0)