I'll show you another way below, but for illustrative purposes,
here's one very slow and tedious way to find out (I used a computer
program to list these, as I wouldn't have had the patience! :)
7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62, 67, 72, 77,
82, 87, 92, 97, 102, 107, 112, 117, 122, 127, 132, 137, 142,
147, 152, 157, 162, 167, 172, 177, 182, 187, 192, 197, 202
As you see the answer is NO. 197 is a term and the next one
is 202, skipping right over 200. But there is another way to find
out without writing them out as the computer did in the list
above.
One better way would be to reason out from the terms you found
7, 12, 17, 22, 27, 32, 37, that they must all either end in
the digit 7 or 2, and since 200 does not end in either of
those digits, it cannot be a term.
But the general way that will work in all cases is to use
this formula for the nth term of an arithmetic sequence:
an = a1 + (n-1)d
The difference between terms is 5 so d=5.
a1 is the first term, which equals 7.
To see if 200 is a term in the sequence we assume
that 200 is a term by substituting it in the formula
above for the nth term an=200, and a1=7, d=5.
200 = (7) + (n-1)(5)
200 = 7 + 5(n-1)
200 = 7 + 5n - 5
200 = 2 + 5n
198 = 5n
= n
= n
But n can only be a whole number, so we have reached a
contradiction, and now we know that 200 will not be a
term in the sequence.
Edwin
