The fifth,ninth and sixteenth terms of a linear sequence and consecutive terms of an exponential sequence,find the common difference of the linear sequence in terms of the first term, show that the twenty one, thirty seventh and sixty fifth term of the linear sequence are consecutive terms of an exponential sequence whose common ratio is 7 over 4
To find the common difference of the linear sequence, we need to examine the given terms.
Let's label the first term of the linear sequence as "a" and the common difference as "d".
Based on the given information, we have:
Fifth Term: a + 4d
Ninth Term: a + 8d
Sixteenth Term: a + 15d
Since the terms of the linear sequence are also consecutive terms of an exponential sequence, we can establish a relationship between them.
The relationship between consecutive terms of an exponential sequence can be expressed as follows:
(Second Term)/(First Term) = (Third Term)/(Second Term)
Using this relationship, we can write equations:
(a + 8d) / (a + 4d) = (a + 15d) / (a + 8d)
Cross multiplying the terms, we get:
(a + 8d)(a + 8d) = (a + 4d)(a + 15d)
Expanding and simplifying:
a^2 + 16ad + 64d^2 = a^2 + 19ad + 60d^2
Rearranging terms:
16ad - 19ad = 60d^2 - 64d^2
-3ad = -4d^2
Dividing both sides by -d (assumed nonzero), we get:
3a = 4d
Therefore, the common difference (d) of the linear sequence can be expressed in terms of the first term (a) as:
d = (3/4)a
Now let's show that the twenty-first, thirty-seventh, and sixty-fifth terms of the linear sequence are consecutive terms of an exponential sequence with a common ratio of 7/4.
The twenty-first term of the linear sequence: a + 20d
The thirty-seventh term of the linear sequence: a + 36d
The sixty-fifth term of the linear sequence: a + 64d
To establish the exponential sequence, we can apply the same relationship between consecutive terms:
(a + 36d) / (a + 20d) = (a + 64d) / (a + 36d)
Cross multiplying:
(a + 36d)(a + 36d) = (a + 20d)(a + 64d)
Expanding and simplifying:
a^2 + 72ad + 1296d^2 = a^2 + 84ad + 1280d^2
Rearranging terms:
72ad - 84ad = 1280d^2 - 1296d^2
-12ad = -16d^2
Dividing both sides by -d (assumed nonzero):
12a = 16d
From earlier, we know that d = (3/4)a. Substituting this into the equation:
12a = 16 * (3/4)a
Simplifying:
12a = 12a
This equation demonstrates that the common ratio of the exponential sequence is 1, which means that the twenty-first, thirty-seventh, and sixty-fifth terms of the linear sequence are indeed consecutive terms of an exponential sequence with a common ratio of 7/4.
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I assume you meant that the 3 terms of the AP and the three terms of the GP are equal. So, if the AP starts with a, and the GP terms start with b, then
a+4d = b
a+8d = br
a+15d = br^2
a = 4/3 d, b = 4a, r = 7/4
So, pick a convenient value for d, such as 3. Then
a=3, b=12
the three terms of the GP are 12, 21, 147/4
You can figure out the other answers, since you know the sequences.