Oh, exponential sequences! Let's get mathematical, and try not to exponentiate too much brainpower.
To find the 3rd term, we can use the formula for exponential sequences, which is generally represented as \(a_n = a_1 \times r^{(n-1)}\), where \(a_n\) is the nth term, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.
Now, you mentioned that the 2nd term is 35 and the 5th term is 4. Let's assume that the first term \(a_1\) is unknown, and we'll call it \(x\). The common ratio \(r\) can be found by dividing any term by the previous term.
For the 2nd term being 35 and the 5th term being 4, \(\frac{{a_2}}{{a_1}} = \frac{{35}}{{x}}\) and \(\frac{{a_5}}{{a_4}} = \frac{{4}}{{35}}\).
Now, let's solve for \(x\) by setting up a proportion using the information we have:
\(\frac{{a_2}}{{a_1}} = \frac{{35}}{{x}}\)
\(\frac{{a_5}}{{a_4}} = \frac{{4}}{{35}}\)
Cross-multiplying gives us:
\(a_2 \times x = a_1 \times 35\) and \(a_5 \times a_4 = 4 \times 35\)
Simplifying further:
\(35x = 35a_1\) and \(a_5 \times a_4 = 140\)
Since we have \(a_5 = 4\) and \(a_1 = x\), we can substitute these values into the equations:
\(35 \times 4 = 35 \times x\)
\(140 = 35x\)
Dividing both sides by 35:
\(x = 4\)
So, the first term \(a_1\) is 4.
Now we can find the 3rd term by substituting the values into the formula:
\(a_3 = a_1 \times r^{(3-1)}\)
\(a_3 = 4 \times r^2\)
Since we don't know the value of \(r\) yet, we need more information to find it. Do you happen to have any additional information, or should we call upon the math fairies for assistance?