The 5th term of an exponential sequence is 4 and of 2nd term is 35(¡)find the 3rd term (¡¡)the 6th term (¡¡¡)the sum of the first 5th term?
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?
To find the 3rd term of the exponential sequence, we first need to identify the common ratio (r) of the sequence. In an exponential sequence, the formula for the nth term is given by:
an = a1 * r^(n-1)
Where an is the nth term, a1 is the first term, r is the common ratio, and n is the position of the term.
Given that the 5th term is 4, we can substitute n = 5 and an = 4 into the formula and solve for a1 * r^4:
4 = a1 * r^4 ---(1)
Next, we are given that the 2nd term is 35. Substituting n = 2 and an = 35 into the formula, we get:
35 = a1 * r^1 ---(2)
Now we have a system of two equations (1) and (2) with two variables (a1 and r). To solve this system, divide equation (1) by equation (2):
(4 / 35) = (a1 * r^4) / (a1 * r) ---> Simplify and rearrange
(4 / 35) = r^3
We can solve for r by taking the cube root of both sides:
r = ∛(4 / 35)
Using a calculator, we find that r is approximately 0.548.
Now that we know the value of r, we can substitute it into equation (2) to solve for a1:
35 = a1 * (0.548)^1
a1 = 35 / 0.548
Once you calculate a1 to be approximately 63.94, you can use the formula to find the 3rd term and the 6th term.
The 3rd term (a3) can be found by substituting n = 3 into the formula:
a3 = a1 * r^(3-1)
Similarly, the 6th term (a6) can be found by substituting n = 6 into the formula:
a6 = a1 * r^(6-1)
Finally, to find the sum of the first five terms, we can use the formula for the sum of an exponential sequence:
Sn = a1 * (r^n - 1) / (r - 1)
Substitute a1, r, and n = 5 into the formula to get the sum of the first five terms (S5).