A series of book was published at 7 year intervals.When the 7th book was issued the sum of publication year is 13524.When was the 1st book published ?

• sol:
  sum of A.P. series=(n/2)(2*a+(n-1)*d)
  here sum=13524,n=7,d=7 substituting these values in above eq. n solving we get
  a=1911
  which was the 1st book published year.
• 12 years agoHelpfull: Yes(79) No(5)
• x+x+7+x+14+x+21+x+28+x+35+x+42=13524
  x=1911
• 12 years agoHelpfull: Yes(39) No(3)
• yes.dis is simple AP series ans is 1911
• 12 years agoHelpfull: Yes(10) No(2)
• 1911,13542/7 =1932
  1932 is d year when 4th book published
  1st book published in year 1911
• 12 years agoHelpfull: Yes(9) No(4)
• take 7 position
  _ _ _ _ _ _ _
  mid one is 13524/7=1932
  1st one should be 1932-21=1911
• 10 years agoHelpfull: Yes(5) No(0)
• 1911...simple AP SERIES
• 12 years agoHelpfull: Yes(2) No(4)
• it is given that series at 7 intervals,
  therefor starting with ,assume 'x'
  x+(x+7)+(x+14)+(x+21)+(x+28)+(x+35)+(x+42)=13,524
  hence x=1911
• 9 years agoHelpfull: Yes(2) No(1)
• this is a dummy question . we get more than 1 answer !!!
• 9 years agoHelpfull: Yes(1) No(8)
• let us sum of the numbers formula,
  let a be the 1st year,
  7yrs is the difference b/n book to book.so,common difference is 7.
  sum of the numbers=13524
  so,Sn=n/2(a+(n-1)d) formula
  13524=(7/2)(a+6*7), by solving this
  a=3822
• 12 years agoHelpfull: Yes(0) No(22)
• the A.P series is like this,
  x+(x+7)+(x+14)+(x+21)+(x+28)+(x+35)+(x+42)=13,524
  simply use the formula of Sn , Sn=(n/2)*(a+l) ..... [here,a=first term =x,l=last term=x+42]
  by solving this we will get x=1911(ANS)
• 7 years agoHelpfull: Yes(0) No(0)
• sol:
  Sn=(n/2)(2*a+(n-1)*d)
  here sum=13524,n=7,d=7.therefore after solving we get-
  a=1911
• 3 years agoHelpfull: Yes(0) No(0)
• A series of books was published at 7 year interval. So the difference among each book is same, i.e., we solve it by A.P series.
  Sum of A.P = n/2(2*a+(n-1)*d)
  by solving we get a=1911
• 3 years agoHelpfull: Yes(0) No(0)