What is the first term of the arithmetic sequence $3,7,11, \ldots$ that exceeds $200$?

Question

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?

Answer 1

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?

Answer 2

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 $199$).

Answer 3

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, $T_{51}=3+4(51-1)=203$