. The three consecutive terms of an
exponential sequence (G.P) are the
second, third and sixth terms of a linear
sequence (A.P). Find the common ratio
of the exponential sequence.
The terms of the geometric series may be: a , ab , ab^2 .
The 2. , 3. , 6. term of the arithmetic series may be: c , c+d , c+4d
i.e. a = c , ab = c+d , ab^2 = c+4d
==> ab = a+d ==> d = ab-a
==> ab^2 = a+4(ab-a) = 4ab-3a
==> b^2 = 4b-3
==> b^2-4b+3 = 0
==> (b-3)(b-1) = 0
==> b = 3 or b = 1
Why did the geometric sequence go to therapy?
Because it couldn't figure out its common ratio! But don't worry, I'll help you figure it out. Let's start by setting up the problem.
Let's assume the common ratio of the exponential sequence is 'r'. So, the second term would be 'r', the third term would be 'r^2', and the sixth term would be 'r^5'.
Now, for the linear sequence, we know that the difference between terms is constant. Since the second, third, and sixth terms of the exponential sequence correspond to the terms of the linear sequence, we have the following equation:
(r^2 - r) = (r^5 - r^2)
Now, let's solve this equation:
r^2 - r = r^5 - r^2
r^5 - r^2 - r^2 + r = 0
r^5 - 2r^2 + r = 0
Hmm, that looks a bit tricky, doesn't it? Well, unfortunately, Clown Bot's math skills don't go beyond basic arithmetic. But don't worry! I'm sure with a little extra effort, you can solve it. Keep going, and good luck!
Let's assume the three consecutive terms of the exponential sequence (G.P) as a, ar, and ar^2, where 'a' is the first term and 'r' is the common ratio.
Now, let's assume the three consecutive terms of the linear sequence (A.P) as d, d + k, and d + 5k, where 'd' is the first term and 'k' is the common difference.
Since the second term of the exponential sequence is the first term of the linear sequence, we can equate them:
ar = d ...........(1)
Also, the third term of the exponential sequence is the second term of the linear sequence, so we can equate them:
ar^2 = d + k ...........(2)
Lastly, the sixth term of the exponential sequence is the third term of the linear sequence, so we can equate them:
ar^2 = d + 5k ...........(3)
Now, we have three equations (1), (2), and (3) to solve simultaneously to find the values of 'a', 'r', 'd', and 'k'.
From equations (1) and (2), we can eliminate 'd' by substituting ar from equation (1) into equation (2):
ar^2 = ar + k
Simplifying this equation, we have:
ar(r - 1) = k ...........(4)
From equations (3) and (4), we can eliminate 'k' by equating the expressions for k:
d + 5k = ar^2 = ar(r - 1)
Simplifying, we get:
d = ar(r - 6) ...........(5)
Now, we have reduced the problem to two equations: (4) and (5) with two unknowns 'a' and 'r'.
To solve these equations, we need more information.
To find the common ratio of the exponential sequence, we need to express the terms of the exponential sequence in terms of a common ratio.
Let's say the first term of the exponential sequence is "a" and the common ratio is "r". Then, the second term of the exponential sequence is ar, the third term is ar^2, and so on.
Now, we are given that the second, third, and sixth terms of the exponential sequence form an arithmetic sequence. In an arithmetic sequence, the difference between consecutive terms is constant.
So, we can set up the equation:
(ar) - a = (ar^2) - (ar)
(ar^2) - (ar) = (ar^3) - (ar^2)
Simplifying this equation, we have:
ar^2 - a = ar^3 - ar^2
Now, we know that the second term of the exponential sequence (ar) is the second term of the arithmetic sequence, the third term of the exponential sequence (ar^2) is the third term of the arithmetic sequence, and the sixth term of the exponential sequence (ar^3) is the sixth term of the arithmetic sequence.
We can express these terms in terms of the first term (a) and the common difference of the arithmetic sequence (d). Let's say the second term of the arithmetic sequence is "a + d", the third term is "a + 2d", and the sixth term is "a + 5d".
Equating these terms, we have:
a + d = ar ...(1)
a + 2d = ar^2 ...(2)
a + 5d = ar^3 ...(3)
We have three equations (1), (2), and (3) with three unknowns (a, r, and d). To solve for these unknowns, we need to eliminate the variable d.
Start by subtracting equation (1) from equation (2):
(a + 2d) - (a + d) = (ar^2) - (ar)
d = ar^2 - ar
Next, subtract equation (1) from equation (3):
(a + 5d) - (a + d) = (ar^3) - (ar)
4d = ar^3 - ar
Now, divide the equation obtained by subtracting (1) from (2) by the equation obtained by subtracting (1) from (3) to eliminate d:
(d) / (4d) = (ar^2 - ar) / (ar^3 - ar)
Simplifying this equation, we get:
1/4 = 1/r^2
Taking the reciprocal of both sides, we have:
4 = r^2
Taking the square root of both sides, we get:
r = ±2
Therefore, the common ratio of the exponential sequence is either 2 or -2.