Latest Elitmus Aptitude Question SOLUTION: 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

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(#M40164963) ELITMUS Numerical Ability question Keep an EYE

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

Asked In Elitmus Satyajit Pramanik (7 years ago)

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pranay kumar  (7 years ago)

8(1,5,7,25,35,125,175,875)

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ehsan  (7 years ago)

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

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yash sharma  (7 years ago)

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

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