What is the first term of the arithmetic sequence $3,7,11, \ldots$ that exceeds $200$? 
The first three terms of a sequence are $3,7,11$. What is the first term to exceed 200?

Here's what I've done so far:
Common difference: $T_2-T_1 = 7-3=4.$
$200 + 4=204$ 
Therefore $204$ is the first term to exceed $200$. 
But when I substitute:
$204=3+(n-1)4$
$204=3+4n-4$
$204=-1+4n$
$204+1=4n$
$205/4=n$
$n=51.25$
It doesn't seem right! What am I doing wrong?
 A: We've already derived the (correct) formula $$T_n := 3 + 4 (n - 1)$$ for the $n$th term of the sequence, and we want to find the smallest $n$ for which $T_n > 200$. Substituting our formula and rearranging gives the (equivalent) inequality
$$n > \tfrac{201}{4}.$$
What is the smallest integer $n$ that satisfies this?
A: If $3,7,11$ are first 3 elements of an arithmetic sequence, then the sequence is $3+4(n-1)$.
You want the smallest $n$ such that $3+4(n-1)>200$, or $4n>201$, or $n>50.25$, so $n=51$ is the smallest one, the 51-th element of the sequence is $203$ (the previous one is $199$).
A: Your approach is correct, but you've miscalculated what the first term over $200$ will be. The first term over $200$ will actually be $203$. To help convince yourself of this, realize that all of the terms in the sequence must be odd.
Your sequence can be written as $T_n=3+4(n-1)$.
Since we wish to determine the term that exceeds $200$, we write $T_n=3+4(n-1)>200$
Back solving, we obtain $n=50.25$. Rounding to account for whole number terms, we get $n=51.$ 
Finally, $T_{51}=3+4(51-1)=203$
