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Chapter 3: Problem 7

Let the sequence \(a_{1}, a_{2}, a_{3}, \ldots\) be in G.P. If the area bounded by the parabolas \(\mathrm{y}^{2}=4 \mathrm{a}_{\mathrm{n}} \mathrm{x}\) and

Short Answer

Expert verified
#Answer# \(a_3 = \left(\frac{96}{x_2^3} + 1\right)^2\)

Step by step solution

01

Find the intersection points of the two parabolas

We are given the parabolas \(y^2 = 4a_nx\) and \(y^2 = 4a_{n+1}x\). We want to find the points where they intersect. To do this, let's set the two equations equal to each other: \(4a_nx = 4a_{n+1}x\) Now, we can divide both sides of the equation by \(4x\): \(a_n = a_{n+1}\) Since the sequence is in G.P., we know that \(a_{n+1} = a_nr\), where \(r\) is the common ratio. So, we can rewrite the equation as: \(a_n = a_nr\) The two parabolas intersect when \(a_n = 0\).
02

Use integration to find the area bounded by the parabolas

The first parabola is given by \(y^2 = 4a_nx\), so \(y = 2\sqrt{a_nx}\). Similarly, the second parabola is given by \(y^2 = 4a_{n+1}x\), so \(y = 2\sqrt{a_{n+1}x}\). The area between these two parabolas is found by integrating the difference between their equations. That is: \(A = \int_{0}^{x_2} (2\sqrt{a_{n+1}x} - 2\sqrt{a_nx}) dx\) We know from the problem statement that this area is \(32\). So: \(32 = \int_{0}^{x_2} (2\sqrt{a_{n+1}x} - 2\sqrt{a_nx}) dx\) Now, we need to solve this integral to find the value of x.
03

Solve the integral

Let's find the indefinite integral first: \(\int (2\sqrt{a_{n+1}x} - 2\sqrt{a_nx}) dx = \frac{4}{3}\sqrt{a_{n+1}x^3} - \frac{4}{3}\sqrt{a_nx^3} + C\) Now, let's integrate over the interval \([0, x_2]\): \(32 = \left[\frac{4}{3}\sqrt{a_{n+1}x^3} - \frac{4}{3}\sqrt{a_nx^3}\right]_0^{x_2}\) This simplifies to: \(32 = \frac{4}{3}(\sqrt{a_{n+1}x_2^3} - \sqrt{a_nx_2^3})\) Now solving for \(\frac{a_{n+1}}{a_n}\) we get: \(\frac{a_{n+1}}{a_n} = r = \frac{96}{x_2^3} + 1\)
04

Calculate the third term of the G.P.

We have found the common ratio \(r\), and we now need to find the third term, which is given by \(a_3 = a_2r\). Since the sequence is in G.P., we have \(a_2 = a_1r\), and therefore, \(a_3 = a_1r^2\). We have the ratio \(r = \frac{96}{x_2^3} + 1\), but we don't know the value of \(a_1\). However, we can see that any multiple of \(a_1\) will result in a valid G.P. with the same common ratio. Therefore, without loss of generality, we can set \(a_1\) to \(1\). Thus: \(a_3 = r^2 = \left(\frac{96}{x_2^3} + 1\right)^2\) Since we don't have the value of \(x_2\), we leave the expression for \(a_3\) in this form.

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