Show that the twenty-first,thirty-seventh and sixty-fifth term of a linear sequnce are consecutive terms of an exponential sequnce whose common ratio is 7/4
To show that the twenty-first, thirty-seventh, and sixty-fifth terms of a linear sequence are consecutive terms of an exponential sequence with a common ratio of 7/4, we need to verify that the ratio of consecutive terms in the linear sequence matches the common ratio in the exponential sequence.
Let's start by finding the general formula for the linear sequence.
In a linear sequence, the nth term (Tn) is given by the formula:
Tn = a + (n - 1)d
Where:
a is the first term
d is the common difference
n is the position of the term in the sequence
Since we don't know the first term (a) and the common difference (d), let's find them using the given information.
First, let's consider the twenty-first term, denoted as T21.
T21 = a + (21 - 1)d
Next, let's consider the thirty-seventh term, denoted as T37.
T37 = a + (37 - 1)d
Finally, let's consider the sixty-fifth term, denoted as T65.
T65 = a + (65 - 1)d
Now, let's set up the ratios using the given exponential sequence common ratio of 7/4:
The ratio of the thirty-seventh term to the twenty-first term would be:
T37 / T21 = (a + 36d) / (a + 20d)
Similarly, the ratio of the sixty-fifth term to the thirty-seventh term would be:
T65 / T37 = (a + 64d) / (a + 36d)
Now, we want to verify if these ratios match the common ratio of the exponential sequence, which is 7/4. So, let's equate these ratios to 7/4:
(a + 36d)/(a + 20d) = 7/4 ...(1)
(a + 64d) / (a + 36d) = 7/4 ...(2)
Now, by solving these two equations simultaneously, we can find the values of a and d. Once we have the values of a and d, we can calculate the terms T21, T37, and T65 to verify if they are consecutive terms of the exponential sequence with a common ratio of 7/4.
To solve equations (1) and (2), we can cross-multiply and simplify:
For equation (1):
4(a + 36d) = 7(a + 20d)
4a + 144d = 7a + 140d
144d - 140d = 7a - 4a
4d = 3a
For equation (2):
4(a + 64d) = 7(a + 36d)
4a + 256d = 7a + 252d
256d - 252d = 7a - 4a
4d = 3a
We can observe that both equations (1) and (2) lead to the same equation:
4d = 3a
This implies that there are infinitely many solutions to this system of equations, as we have one equation in two variables.
Therefore, it is not possible to uniquely determine the values of 'a' and 'd' to show that the twenty-first, thirty-seventh, and sixty-fifth terms of a linear sequence are consecutive terms of an exponential sequence with a common ratio of 7/4.
(a+36d)/(a+20d) = 7/4
(a+64d)/(a+36d) = 7/4
3a=4d
looks like there are many such sequences:
a=12, d=9
T21 = 192
T37 = 336
T65 = 588