.A series of books was published at 10years intervals when the 10th book was issued the sum of publication years was 19,560when was the 1st book published
a)1910 b)1914 c)1911 d)1909

• ans is 1911
  here d=10
  sn=19560
  n=10
  we have to find a
  so sn=n/2(2a+(n-1)d)
  now put values
  19560=10/2(2a+9*10)
  after solving we get a=1911
  
• 9 years agoHelpfull: Yes(4) No(1)
• 19560=10/2[a+a+(10-1)*10]
  19560=5[2a+90]
  19560-450=10a
  a=19110/10
  a=1911
• 9 years agoHelpfull: Yes(3) No(1)
• 19560 is the answer
• 8 years agoHelpfull: Yes(0) No(10)
• Answer is 1911
• 4 years agoHelpfull: Yes(0) No(0)