if any set of numbers starting with {2,3,4,.....876,877} . how many A.P. can be formed if every A.p. contains at least 3 terms and starting with 2 and ending with 877.
a)-6 b)-7 c)-9 d)-12

• 2+ (n-1)k=877 i.e we need factors of 875 which are 8. but k=875 cant be taken , hence answer is 7
• 9 years agoHelpfull: Yes(16) No(3)
• hey guys the ans is 6....
  we have first term as 2 and last term is 877
  now according to last term formula of an AP i.e., a+(n-1)d
  we have (n-1)d=875
  now 875 is divided by 1,5,7,35,125,175
  it means if we add these nos with 2 we can get 877 at last
  hence we can generate 6 different Ap strating from 2 to 877
  thank u ....
• 9 years agoHelpfull: Yes(4) No(3)
• total no of digit is 876
  prime factor of 876=2^2*3*73
  total no of divisor will be 3*2*2=12 ans
  (no of divisor is found by the multiplication of degree+1)
• 9 years agoHelpfull: Yes(1) No(3)
• answer cant be negative
• 9 years agoHelpfull: Yes(0) No(0)
• 7 bcoz 2+(n-1)*d=877
  (n-1)*d=875
  factor of 875 is 8 but we cannot include d=875 so ans-7
  
• 8 years agoHelpfull: Yes(0) No(0)