Chapter 18: Problem 6
Find the sum of the three numbers in G.P. whose product is 216 and the sum of the products of them taken in pairs is \(126:\) (a) 28 (b) 21 (c) \(35 / 4\) (d) none of these
Short Answer
Expert verified
Answer: (c) 35/4
Step by step solution
01
Write down ABC = 216
Using the notation defined above, we have:
(ar^2) * (ar) * (ar^(-2)) = 216
02
Simplify the equation
We can simplify the above equation by simplifying the exponents:
a^3 = 216
03
Find the value of a
From the above equation, we can find the value of a:
a = 6
04
Write down AB + AC + BC = 126
Using the notation defined above, we have:
(6r^2) * (6r) + (6r^2) * (6r^(-2)) + (6r) * (6r^(-2)) = 126
05
Simplify the equation
We can simplify the above equation by simplifying the terms:
36r^3 + 36 + 36/r = 126
06
Move the constant term to the other side
We can move the constant term to the right side of the equation:
36r^3 + 36/r = 90
07
Multiply the equation by r
To get rid of the term with r in the denominator, we can multiply the whole equation by r:
36r^4 + 36 = 90r
08
Move the terms involving r to the left side
We can move the terms involving r to the left side of the equation:
36r^4 - 90r + 36 = 0
09
Factor the equation
This equation can be factored as follows:
(2r - 3)(18r^3 + 6r^2 - 12r + 12) = 0
10
Find the value of r
The first factor, (2r - 3), tells us that the value of r should be:
r = 3/2
11
Calculate A, B, and C
Using the value of a (6) and the value of r (3/2), we can find the three numbers in G.P.:
A = 6 * (3/2)^2 = 81/4
B = 6 * (3/2) = 9
C = 6 * (3/2)^(-2) = 16/3
12
Find the sum A + B + C
The sum of the three numbers A, B, and C is:
A + B + C = 81/4 + 9 + 16/3
A + B + C = 35/4
So the sum of the three numbers is 35/4, which corresponds to choice (c).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometric Progression Sum
Understanding the sum of a geometric progression (GP) is essential for solving many mathematical problems. A geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed non-zero number called the common ratio (r). For instance, in the sequence 2, 4, 8, 16..., the common ratio is 2 because each term is twice the preceding term.
The sum of a finite GP is calculated using the formula:
\[ S_n = \frac{a(1-r^n)}{1-r} \]
where \( a \) is the first term, \( r \) is the common ratio, \( n \) is the number of terms, and \( S_n \) is the sum of the first \( n \) terms. However, if you're dealing with an infinite GP, and the common ratio's absolute value is less than 1, the sum can be found using the formula:
\[ S = \frac{a}{1-r} \]
In our exercise, the problem hints at a GP involving three terms with a known product and a sum of products taken in pairs. Such nuances require a clear understanding of the properties of GPs to solve efficiently.
The sum of a finite GP is calculated using the formula:
\[ S_n = \frac{a(1-r^n)}{1-r} \]
where \( a \) is the first term, \( r \) is the common ratio, \( n \) is the number of terms, and \( S_n \) is the sum of the first \( n \) terms. However, if you're dealing with an infinite GP, and the common ratio's absolute value is less than 1, the sum can be found using the formula:
\[ S = \frac{a}{1-r} \]
In our exercise, the problem hints at a GP involving three terms with a known product and a sum of products taken in pairs. Such nuances require a clear understanding of the properties of GPs to solve efficiently.
Geometric Sequences
Geometric sequences are a list of numbers where each term after the first is found by multiplying the previous term by a constant called the ratio. It is a foundational concept within arithmetic and widely encountered in various mathematical contexts.
Each term in a geometric sequence can be found using the formula:
\[ a_n = a_1 \times r^{(n-1)} \]
where \( a_n \) is the nth term, \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term's position in the sequence. Knowing how to manipulate the properties of geometric sequences is crucial for solving problems that ask for specific terms, their sum, or the overall pattern they describe.
In our textbook exercise, we identify three terms of a GP. To find these terms, we need to establish the first term (a) and the common ratio (r), which then allow us to determine any term in the sequence.
Each term in a geometric sequence can be found using the formula:
\[ a_n = a_1 \times r^{(n-1)} \]
where \( a_n \) is the nth term, \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term's position in the sequence. Knowing how to manipulate the properties of geometric sequences is crucial for solving problems that ask for specific terms, their sum, or the overall pattern they describe.
In our textbook exercise, we identify three terms of a GP. To find these terms, we need to establish the first term (a) and the common ratio (r), which then allow us to determine any term in the sequence.
Product of GP Terms
The product of terms in a geometric progression has its own unique properties and understanding how to calculate it is key for tackling GP-related problems. In a sequence of a GP with terms \(a_1, a_2, a_3, ..., a_n\), the product of the first \(n\) terms can be expressed as:
\[ P_n = (a_1 \times a_2 \times ... \times a_n) = a^n \times r^{\frac{n(n-1)}{2}} \]
where \(P_n\) is the product of the first \(n\) terms, \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms. If we take the nth root of this product, we get back to the first term \(a\).
In the exercise, we used the knowledge of the product of the GP terms to first establish the value of \(a^3\), which then helped us in deducing the value of \(a\). From there, it was a matter of using the found values to calculate the sum of the sequence.
\[ P_n = (a_1 \times a_2 \times ... \times a_n) = a^n \times r^{\frac{n(n-1)}{2}} \]
where \(P_n\) is the product of the first \(n\) terms, \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms. If we take the nth root of this product, we get back to the first term \(a\).
In the exercise, we used the knowledge of the product of the GP terms to first establish the value of \(a^3\), which then helped us in deducing the value of \(a\). From there, it was a matter of using the found values to calculate the sum of the sequence.
