how many a.p series are possible where Ist digit is 2 and the last digit is 877 that contains at least 3numbers in series

• 8(1,5,7,25,35,125,175,875)
• 6 years agoHelpfull: Yes(8) No(2)
• l=a+(n-1)d
  877=2+(n-1)d
  875/d=n-1
  875/m=d suppose m=n-1 now 875 is divisible by m so we can count number
  last digit is 5 so id 5
  875/5=175/5=35/5=7 we can see 875 is divisible by {5,25,125}
  again last is 7 that means it is also divisible by 7
  875/7=125/5=25/5=5 875 is divisible by {7,35,175}
  total={1,5,25,125,7,35,175} 7 number
• 6 years agoHelpfull: Yes(8) No(1)
• given a=2 l=875
  l=a+(n-1)d
  877=2+(n-1)d
  if d is integer and n is integer than
  875/d=n-1
  n is integer if d is 1,5,7,25,25
  so 5 aps are possible
• 6 years agoHelpfull: Yes(0) No(6)