Finding the largest geometric term below a threshold

Expert intro: The geometric sequence 448, 224, 112, ... uses a common ratio of one-half, so the size of each term can be tracked with a power expression [1][2].

Detailed walkthrough

Identify the pattern in the geometric sequence

The sequence 448, 224, 112, ... has first term $a_1 = 448$ and common ratio $r = \tfrac{1}{2}$. That means the nth term is

$a_n = 448\left(\tfrac{1}{2}\right)^{n-1}.$

Because the terms are decreasing, the “largest term smaller than 2” is the last term that is still below 2.

We therefore want the smallest positive integer $n$ such that

$448\left(\tfrac{1}{2}\right)^{n-1} < 2.$

Once we find that first index below 2, the required answer is simply that term itself.

Solve the inequality for the index

The expression $448\left(\tfrac{1}{2}\right)^{n-1} < 2$ can be rewritten by dividing both sides by 448:

$\left(\tfrac{1}{2}\right)^{n-1} < \tfrac{1}{224}.$

Taking logarithms gives

$(n-1)\log\left(\tfrac{1}{2}\right) < \log\left(\tfrac{1}{224}\right).$

Since $\log\left(\tfrac{1}{2}\right)$ is negative, the inequality reverses when we divide by it:

$n-1 > \frac{\log(1/224)}{\log(1/2)}.$

Numerically, this is a little less than 8.80, so the smallest integer satisfying the condition is $n=9$.

That means the first term smaller than 2 is the 9th term.

Compute the required term and check it

Using the formula,

$a_9 = 448\left(\tfrac{1}{2}\right)^8 = \frac{448}{256} = \frac{7}{4}.$

This is indeed smaller than 2, and the previous term is

$a_8 = 448\left(\tfrac{1}{2}\right)^7 = \frac{7}{2},$

which is larger than 2. So $\frac{7}{4}$ is the largest term in the sequence that is still below 2.

A quick pattern check is useful here: because each term is exactly half the previous term, the crossing point must happen between two consecutive terms, and the inequality method confirms the exact index.

💡 Pitfall guide

A common mistake with 448, 224, 112, ... is to stop at the first term that is close to 2 instead of checking the first term that is actually below 2. Since the sequence decreases by a factor of one-half each time, the boundary matters: 7/2 is still above 2, while 7/4 is below 2. Another trap is mishandling logarithms when the base ratio is less than 1. Because log(1/2) is negative, the inequality direction changes when you divide by it. If you forget that sign flip, you will get the wrong index and may choose the 8th term instead of the 9th. It also helps to verify the answer directly by substituting consecutive terms rather than trusting only the algebra.

🔄 Real-world variant

If the sequence were 448, 224, 112, ... and the threshold changed from 2 to 5, the same method would still work, but the cutoff index would change. You would solve 448(1/2)^(n-1) < 5, find the first integer n that makes the term drop below 5, and then test the neighboring term above it. A second variant is to keep the threshold at 2 but change the common ratio to 3/4, giving 448(3/4)^(n-1) < 2. That version requires the same inequality setup, but the decay is slower, so the index would be much larger. In either variation, the key idea is to identify the nth-term formula first, then locate the first term that crosses the boundary.

🔍 Related terms

geometric sequence nth term, common ratio inequality, largest term below threshold