In the sequence 14, 18, 22, 26, …, every term is four more than the previous term. How many of the first 100 terms are divisible by

Question

In the sequence 14, 18, 22, 26, …, every term is four more than the previous term.
How many of the first 100 terms are divisible by 8? ANSWER IT PLEASE!!!!!!!!!!!!

in progress 0
Mary 3 weeks 2021-10-01T20:37:12+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-10-01T20:38:46+00:00

    general term for sequence is

    10 + 4n where n = 1,2,3,…

    4n + 10 = 2(mod 4)

    we have to find n for which

    4n + 10 = 0 (mod 8)

    4n = -10 (mod 8)

    4n = 6 ( mod 8)

    here gcd of 4n and 8 is atleast 4.

    4 does not divide 6

    hence, this congruence has no solution

    i.e. there is no term in the sequence which will be divisible by 8.

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )