Show that the 21st, 37th and 65th terms of the linear sequence are consecutive terms of an exponential sequence whose common ratio is 7/4.
What kind of website is this
No brain to solve this question😒😑😡😵🤬🧐
To show that the 21st, 37th, and 65th terms of a linear sequence are consecutive terms of an exponential sequence with a common ratio of 7/4, we need to analyze the given information and find a relationship between the terms.
Let's start by considering a linear sequence, where each term can be represented as:
an = a1 + (n-1)d
Here, 'an' represents the nth term, 'a1' is the first term, 'n' is the position of the term, and 'd' represents the common difference between consecutive terms.
Given that the 21st term is the first term, the 37th term is the second term, and the 65th term is the third term, then we can set up the following equations:
a21 = a1 + 20d --- (1)
a37 = a1 + 36d --- (2)
a65 = a1 + 64d --- (3)
Now let's consider an exponential sequence where each term is related to the previous term by a constant ratio, 'r':
an = a1 * r^(n-1)
Here, 'an' represents the nth term, 'a1' is the first term, 'n' is the position of the term, and 'r' represents the common ratio.
Since the 21st, 37th, and 65th terms of the linear sequence are consecutive terms of the exponential sequence, we can write the following relationships:
a21 = a1 * r^20 --- (4)
a37 = a1 * r^36 --- (5)
a65 = a1 * r^64 --- (6)
Our goal is to show that the common ratio, 'r', is equal to 7/4. To do this, we can compare the given equations.
From equations (1) and (4), we have:
a1 + 20d = a1 * r^20
Rearranging the equation, we get:
r^20 = (a1 + 20d) / a1
Similarly, from equations (2) and (5), we have:
r^36 = (a1 + 36d) / a1
And from equations (3) and (6), we have:
r^64 = (a1 + 64d) / a1
Now, let's compare the equations to find a relationship between them.
r^20 = (a1 + 20d) / a1 --- (7)
r^36 = (a1 + 36d) / a1 --- (8)
r^64 = (a1 + 64d) / a1 --- (9)
Taking equation (7) and raising it to the power of 16 (since 20*16 = 320), we get:
(r^20)^16 = [(a1 + 20d) / a1]^16
Simplifying, we have:
r^320 = [(a1 + 20d) / a1]^16
Similarly, taking equation (8) and raising it to the power of 9 (since 36*9 = 324), we get:
(r^36)^9 = [(a1 + 36d) / a1]^9
Simplifying, we have:
r^324 = [(a1 + 36d) / a1]^9
Finally, taking equation (9) and raising it to the power of 5 (since 64*5 = 320), we get:
(r^64)^5 = [(a1 + 64d) / a1]^5
Simplifying, we have:
r^320 = [(a1 + 64d) / a1]^5
Comparing the equations, we can see that r^320 is common in all three equations, and simplifying these equations confirms that:
[(a1 + 20d) / a1]^16 = [(a1 + 36d) / a1]^9 = [(a1 + 64d) / a1]^5
Now, since [(a1 + 20d) / a1]^16 = [(a1 + 36d) / a1]^9, we can write:
[(a1 + 20d) / a1]^16 / [(a1 + 36d) / a1]^9 = 1
Simplifying this expression, we get:
(a1 + 20d)^16 / (a1 + 36d)^9 = 1
Taking the 9th root of both sides, we have:
[a1 + 20d) / (a1 + 36d)]^(16/9) = 1
Since anything raised to the power of zero equals 1, we can say:
[a1 + 20d) / (a1 + 36d)]^(16/9) = [(a1 + 64d) / a1) / [(a1 + 64d) / a1]
Which simplifies to:
[a1 + 20d) / (a1 + 36d)]^(16/9) = 1
This equation holds true when the numerator and denominator are equal to each other:
a1 + 20d = a1 + 36d
Canceling out the common terms, we get:
4d = 0
Since 'd' represents the common difference of a linear sequence, this equation implies that the common difference is zero. However, in a linear sequence, a common difference of zero would mean that all terms in the sequence are equal, which contradicts the fact that terms 21, 37, and 65 are different.
Therefore, our conclusion is that the 21st, 37th, and 65th terms of a linear sequence cannot be consecutive terms of an exponential sequence with a common ratio of 7/4, as there exists no solution that satisfies the given conditions.
which sequence?
I can easily show that in general they are not.