Latest Elitmus Aptitude Question SOLUTION: 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)

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(#M40151469) ELITMUS Numerical Ability question Keep an EYE

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

Asked In Elitmus ankit (9 years ago)

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Kamal Chaturvedi  (9 years ago)

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

Is this solution Helpfull? Yes (16) | No (3)  

sagar chhetri  (9 years ago)

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 ....

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Dhananjay  (9 years ago)

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)

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soni kumari  (9 years ago)

answer cant be negative

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Kumar Anmol   (9 years ago)

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

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