in the sequance a three digit numbers below 500 which leaves riminder 1 when divided by 7 what is the first and last term pls
Answers
Step-by-step explanation:
Given :-
the sequance a three digit numbers below 500 which leaves riminder 1 when divided by 7
To find:-
what is the first and last term ?
Solution:-
The list of three digits numbers =
100,101,102,...,999
The list of three digits numbers below 500 =
100,101,102,....,498,499.
The list of multiples of 7 below 500
=105, 112,119,...,490,497.
The list of numbers which leaves remainder 1 when divided by 7 below 500
=>106,113,...,491,498.
The sequence of the numbers which leaves remainder 1 when divided by 7 below 500
=106, 113 , ..., 491 ,498
In the sequence ,
The first term = 106
The last term = 498
Answer:-
The first term of the sequence = 106
The last term of the sequence = 498
Additional information:-
The general form of the numbers which divided by 7 = 7n
The numbers which leaves remainder 1 when divided by 7 = 7n+1