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Chapter 8: Problem 624

Find out three numbers which are in G. P. such that their summation is 13 and the sum of their squares is 91 (A) \(3,1,9\) (B) \(1,3,9\) (C) \(1,9,3\) (D) \((13 / 3),(13 / 3),(13 / 3)\)

Short Answer

Expert verified
(B) \(1,3,9\)

Step by step solution

01

Write the general form of G.P.

For any G.P. with three terms, we can write the terms as follows: \(a\), \(ar\), and \(ar^2\), where \(a\) is the first term, and \(r\) is the common ratio.
02

Write the given conditions in terms of the first term and the common ratio

We are given that the sum of the terms is 13 and the sum of their squares is 91. Let's write these equations: \(a + ar + ar^2 = 13\) and \(a^2 + a^2r^2 + a^2r^4 = 91\)
03

Simplify the equations and solve for the variables

The first equation can be rewritten as a factored equation: \(a(1 + r + r^2) = 13\) Divide the second equation by \(a^2\) \(1 + r^2 + r^4 = \frac{91}{a^2}\) Now, let's substitute \(\frac{13}{1+r+r^2}\) for \(a\) in the second equation: \(1 + r^2 + r^4 = \frac{91}{(\frac{13}{1+r+r^2})^2}\) Now solve the equation for \(r\). It's a quartic equation, which is quite advanced to solve manually. You could use a quartic equation solver online or a calculator with a polynomial solver. Once you've got the value of \(r\), you can substitute it back to find the value of \(a\). After solving for \(r\) and \(a\), we get: \(a = 1\) \(r = 3\)
04

Determine the numbers in the G.P.

Using the values of \(a\) and \(r\), we can find the three numbers: The first term: \(1\) The second term: \(1*3 = 3\) The third term: \(3*3 = 9\) Hence, the three numbers are \(\boldsymbol{1, 3, 9}\). Comparing our answer with the options, we find that the correct answer is: (B) \(\boldsymbol{1,3,9}\)

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Most popular questions from this chapter

If \((1 / a),(1 / b),(1 / c)\) are in A. P., then \([(1 / a)+(1 / b)-(1 / c)]\) \([(1 / b)+(1 / c)-(1 / a)]=\) (A) \(\left[\left(4 b^{2}-3 a c\right) /(a b c)\right]\) (B) \((4 / \mathrm{ac})-\left(3 / \mathrm{b}^{2}\right)\) (C) \((4 / \mathrm{ac})-\left(5 / \mathrm{b}^{2}\right)\) (D) \(\left[\left(4 b^{2}+3 a c\right) /\left(a b^{2} c\right)\right]\)

The coefficient of \(x^{8}\) in the product \((x+1)(x+2)(x+3) \ldots\) \((\mathrm{x}+10)\) is (A) 1024 (B) 1300 (C) 1320 (D) 1360

Sum of products of first n natural numbers taken two at a time is (A) \(\left[\left\\{\overline{ \left.n\left(n^{2}-1\right)(3 n+2)\right\\} /(24)}\right]\right.\) (B) \(\left[\left\\{n(n+1)^{2}(3 n+2)\right\\} /(72)\right]\) (C) \(\left[\left\\{n^{2}(n+1)(3 n+2)\right\\} /(48)\right]\) (D) \([\\{n(n+1)(n+2)(3 n+2)\\} /(96)]\)

\(\tan ^{-1}(1 / 3)+\tan ^{-1}(1 / 7)+\tan ^{-1}(1 / 13)+\ldots+\tan ^{-1}[1 /(9703)]\) (A) \(\begin{array}{lll}\text { (B) }(\pi / 6) & \text { (C) }(\pi / 3) & \text { (D) } \tan ^{-1}(0.98)\end{array}\)

\((1 / 3)+\left(2 / 3^{2}\right)+\left(1 / 3^{3}\right)+\left(2 / 3^{4}\right)+\left(1 / 3^{5}\right)+\left(2 / 3^{6}\right)+\ldots\) up to \(\infty=\) (A) \((1 / 8)\) (B) \((3 / 8)\) (C) \((7 / 8)\) (D) \((5 / 8)\)
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