The 49th term of an arithmetic progression is 23 and 62nd term is 37. What is the sum of the first 110 terms of the series?
option
a) 3000
b) 3300
c) 3600
d) 3900

• a+48d=23
  a+61d=37
  on solving we get a=-28.69 d=14/13
  now a110=a+109d=88.69
  so sum of 110 term=n/2(a+a110) where n =110 so on solving we get 3300ans
  
• 11 years agoHelpfull: Yes(12) No(0)
• To find sum of 110 numbers
  sum of 110 numbers=(110/2)(a+(n-1)d)
  =55(a+(n-1)d)
  
  so the answer should be a multiple of 55
  3000/55=54.54
  3300/55=60
  3600/55=65.45
  3900/55=70.9
  
  so the ans is b)3300
• 11 years agoHelpfull: Yes(5) No(0)
• a+48d=23
  a+61d=37
  so wo have d=14/13 and
  a=-373/13
  from the formula s(n)=n/2(2a+(n-1)d) =3300
  
• 11 years agoHelpfull: Yes(1) No(0)
• 3295005.....using simple ap
• 11 years agoHelpfull: Yes(0) No(2)