If a publication occurs every seven years and the sum of the years is 13524. Then find the first year

• a + ( a + 7) + (a + 14) + …. Sum of 7 terms = 13524.
  Thus, i.e. 2a + 42 = 3864 i.e. a = 1911
  
• 10 years agoHelpfull: Yes(27) No(6)
• NEED HELP
  If a publication occurs every seven years,
  then d = 7 and given, Sn = 13524
  For those using AP Series, can someone please explain me, from which part of question are you interpreting n = 7
  
• 10 years agoHelpfull: Yes(20) No(2)
• let is term is a.
  a+(a+7)+(a+14)........7terms
  s=n/2(2a+(n-1)d)
  13524=7/2(2a+6*7)
  a=1911
• 10 years agoHelpfull: Yes(14) No(1)
• you have take to the 1911,
  1911-1918 is released 1st book
  1918-1925 is released 2nd book
  1925-1932 is released 3rd book
  1932-1939 is released 4th book
  1939-1946 is released 5th book
  1946-1953 is released 6th book
  1953-1960 is released 7th book.
  So add the years 1911+1918+1925+1932+1939+1946+1953=13,524.
  ANS IS 1911.
• 10 years agoHelpfull: Yes(9) No(11)
• on what basis u r taking n=7.??? statement tells d=7
• 10 years agoHelpfull: Yes(7) No(1)
• let 1 yr be x
  x+6x=13524
  7x=13524
  x=1932;
  6x=11592
  x=13524-11592
  x=1932(ans)
  
• 10 years agoHelpfull: Yes(4) No(6)
• answer is 1929.7years .let 1st year b x . x, x+1, x+2..... 7x+21=13524, 7x=13503, x=1929.
• 10 years agoHelpfull: Yes(2) No(2)
• Ans:1911 is right answer
  we can solve it by AP and sum of ap is n/2(2a+(n-1)d)
  so we have series a+(a+7)+(a+14).......=13524;
  13524=7/2(2*a+(7-1)7)
  a=1911;
• 10 years agoHelpfull: Yes(2) No(1)
• n/2(2a+(n-1)6)=13524
  n is no of publications
• 10 years agoHelpfull: Yes(1) No(4)
• 1932 years back
  13524/7
  ,2013-1932= 181
• 10 years agoHelpfull: Yes(0) No(4)
• sorry d is 7
• 10 years agoHelpfull: Yes(0) No(0)
• we all know this simple formula,Sn=(n/2)(2a+(n-1)d)
  now identify "a".
  13524=(7/2)(2a+(7-1)7)
  after we solve , we get a=1911
• 10 years agoHelpfull: Yes(0) No(4)
• let 1st year is x
  x+7x=13524.
  x=1677
• 10 years agoHelpfull: Yes(0) No(1)
• a + ( a + 7) + (a + 14) + …. Sum of 7 terms = 13524.
  Thus, i.e. 2a + 42 = 3864 i.e. a = 1911
  
  :p
• 10 years agoHelpfull: Yes(0) No(0)
• a+(a+7)+(a+14)+........sum of 7 term = 13524
  
  i.e. n/2 [2a+(n-1)d] = 13524
  7/2[ 2a+(7-1)*7] = 13524
  2a + 42 = 3864
  a =1911
• 8 years agoHelpfull: Yes(0) No(0)
• Try This.............Logically from option given
  you have take to the 1911,
  1911-1918 released 1st book
  1918-1925 released 2nd book
  1925-1932 released 3rd book
  1932-1939 released 4th book
  1939-1946 released 5th book
  1946-1953 released 6th book
  1953-1960 released 7th book.
  So addition done= 1911+1918+1925+1932+1939+1946+1953=13,524.
  Hence ans= 1911.
• 7 years agoHelpfull: Yes(0) No(0)
• 1911 is correct answer. Because other options are earlier years. There were no availability of publications.
• 7 years agoHelpfull: Yes(0) No(0)
• just check in options which are divisible by 7. :-)
• 7 years agoHelpfull: Yes(0) No(0)