Question

The ratio of the sum of the first three terms of a GP to the sum of its first six terms is \(125: 152\). The common ratio of the GP is (a) \(\frac{1}{5}\) (b) \(\frac{2}{5}\) (c) \(\frac{3}{5}\) (d) \(\frac{4}{5}\)

Short answer

Answer: The common ratio of the geometric progression is \(\frac{2}{5}\).

Step-by-step solution

Write general expressions for the sum of the first three and six terms of a GP

Let the first term of the GP be "a" and the common ratio be "r". Then, the sum of the first three terms, S3, and the sum of the first six terms, S6, can be expressed as: S3 = a(1 + r + r^2) S6 = a(1 + r + r^2 + r^3 + r^4 + r^5)

Use the given ratio to form an equation

We are given that the ratio of S3 to S6 is 125:152. Let's write this as a fraction and set up an equation with S3 and S6: \(\frac{S3}{S6} = \frac{125}{152}\) Now substitute the expressions for S3 and S6 from step 1: \(\frac{a(1 + r + r^2)}{a(1 + r + r^2 + r^3 + r^4 + r^5)}= \frac{125}{152}\)

Solve the equation for the common ratio, r

To solve for r, we can cancel out a from both numerator and denominator, and then simplify: \(\frac{1 + r + r^2}{1 + r + r^2 + r^3 + r^4 + r^5} = \frac{125}{152}\) Now cross-multiply and simplify: \((1 + r + r^2)(152) = (1 + r + r^2 + r^3 + r^4 + r^5)(125)\) Expanding both sides and canceling common terms, we get: \(-125r^3 - 125r^4 - 125r^5 = -152r^2\) Divide both sides by -125: \(r^3 + r^4 + r^5 = \frac{152}{125}r^2\) Now, let's factor r^2 from the right side: \(r^2(r + r^2 + r^3) = \frac{152}{125}r^2\) Since r^2 ≠ 0, we can divide both sides by r^2: \(r + r^2 + r^3 = \frac{152}{125}\)

Test the given options for the common ratio, r

Let's test each option (a), (b), (c), and (d) in our equation: (a) r = \(\frac{1}{5}\): \(\frac{1}{5} + \frac{1}{25} + \frac{1}{125} = \frac{156}{125}\) (b) r = \(\frac{2}{5}\): \(\frac{2}{5} + \frac{4}{25} + \frac{8}{125} = \frac{152}{125}\) (c) r = \(\frac{3}{5}\): \(\frac{3}{5} + \frac{9}{25} + \frac{27}{125} = \frac{206}{125}\) (d) r = \(\frac{4}{5}\): \(\frac{4}{5} + \frac{16}{25} + \frac{64}{125} ≠ \frac{152}{125}\)

Identify the correct common ratio

We tested all four options, and only (b) \(\frac{2}{5}\) satisfies our equation. Therefore, the common ratio of the GP is: r = \(\frac{2}{5}\) (Option (b))

Key concepts

Geometric Progression (GP)

A Geometric Progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number known as the common ratio. For example, in the sequence 2, 6, 18, 54, ..., the common ratio is 3 since each term is thrice the previous term. Understanding this multiplication pattern is key to not only recognizing a GP but also to determining the sum of its terms and solving related algebraic equations.

Moreover, the formula for the nth term of a GP (\( a_n \)) if the first term is \( a \) and the common ratio is \( r \) is given by \( a_n = a \times r^{(n-1)} \) where \( n \) is the position of the term in the sequence. Recognizing and utilizing the properties of GPs allow us to solve various mathematical problems involving sequences.

Sum of terms in GP

Knowing how to calculate the sum of terms in a Geometric Progression is crucial, especially in cases where you need to understand the behavior of the GP when summed over a number of terms. The sum of the first \( n \) terms of a GP (\( S_n \) for \( n \) terms) can be given by two main formulas.

If the common ratio \( r \) is not equal to 1, \( S_n \) is calculated as \( S_n = a \times \frac{1 - r^n}{1 - r} \) where \( a \) is the first term.

However, if the common ratio \( r \) is equal to 1, all the terms in the series are the same as \( a \) and the sum \( S_n \) is simply \( n \times a \) . This formula helps simplify problems regarding the total of terms in a GP to a great extent, making it easier to derive and solve related algebraic equations.

Ratio and Proportion

Ratio and proportion are fundamental concepts of mathematics, particularly useful when comparing amounts and solving sequence-related problems. A ratio is a way to compare two quantities by using division, expressed as \( a:b \) or \( \frac{a}{b} \) where \( a \) and \( b \) are the quantities being compared.

A proportion, on the other hand, is an equation that states that two ratios are equivalent. For example, \( \frac{a}{b} = \frac{c}{d} \) is a proportion. When working with series and sequences like GP, the concept of ratio and proportion becomes indispensable as it helps in understanding the relation between the parts of the sequence and the whole. Being adept in manipulating these relationships often simplifies complex problems and is essential for solving algebraic equations involving sequences.

Algebraic Equation Solving

Algebraic equation solving is a systematic process that allows us to find the value of unknowns. In the context of a GP, it often involves finding the common ratio based on the relationships between the terms of the sequence or given conditions. The strategy involves representing the unknowns by variables, forming equations, and then simplifying these equations to solve for the variables.

The step-by-step solution above is a perfect illustration of algebraic problem-solving: It starts with setting up the relation between known and unknown quantities using expressions, proceeds to equation formulation by equating ratios as shown in the problem's condition, and then systematically solves for the unknown common ratio. Mastery of algebraic techniques such as factoring, distributing, and using properties of equality can therefore be very beneficial and are applicable in a wide array of mathematical problems, not limited to just sequences.