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Chapter 22: Problem 702

The sum of an infinite number of terms in geometric progression is 15 , and the sum of their squares is 45 ; find the sequence. Assume that the common ratio of the G.P. is less than 1 .

Short Answer

Expert verified
The infinite geometric progression is \(\frac{15}{2}, \frac{15}{4}, \frac{15}{8}, \frac{15}{16}, ...\), with the common ratio \(r = \frac{1}{2}\).

Step by step solution

01

Sum formula of the infinite geometric progression

Let's denote the first term of the geometric progression as "a" and the common ratio as "r". Since we know that the sum of the infinite geometric progression is 15, we can use the formula for the sum of an infinite geometric progression: \(S_\infty = \frac{a}{1 - r}\) where \(S_\infty\) is the sum of the infinite geometric progression. In this case, we have: \(15 = \frac{a}{1 - r}\)
02

Sum formula of the squares of terms in an infinite geometric progression

Now, let's find the formula for the sum of the squares of the terms in an infinite geometric progression. We can denote this sum as \(S^2_\infty\), which is given by the formula: \(S^2_\infty = \frac{a^2}{1 - r^2}\) Given that the sum of the squares of the terms in our geometric progression is 45, we have: \(45 = \frac{a^2}{1 - r^2}\)
03

Solving the system of equations

We now have a system of two equations with two variables, "a" and "r". We can solve this system by substitution or elimination method. We'll use the substitution method. First, we'll make "a" the subject of the equation in Step 1: \(a = 15(1 - r)\) Now, we'll substitute this expression for "a" into the equation in Step 2: \(45 = \frac{(15(1 - r))^2}{1 - r^2}\)
04

Finding the common ratio r

By solving the equation from Step 3, we'll find the common ratio "r". Simplify the equation: \(45(1 - r^2) = 225(1 - r)^2\) Divide both sides by 15: \(3(1 - r^2) = 15(1 - r)^2\) Now, solve the quadratic equation: \(0 = 32r^2 - 60r + 30\) \(0 = 16r^2 - 30r + 15\) Using the quadratic formula, we can find that: \(r = \frac{1}{2}\)
05

Finding the first term a

Now, we can find the first term "a" of the geometric progression by substituting the value of "r" into the expression for "a" we derived in Step 3: \(a = 15(1 - \frac{1}{2})\) \(a = 15(\frac{1}{2})\) \(a = \frac{15}{2}\)
06

Write the Geometric Progression

Now that we've found the first term "a" and the common ratio "r", we can write the geometric progression: \(\frac{15}{2}, \frac{15}{4}, \frac{15}{8}, \frac{15}{16}, ...\) Thus, the infinite geometric progression is: \(\frac{15}{2}, \frac{15}{4}, \frac{15}{8}, \frac{15}{16}, ...\), with the common ratio \(r = \frac{1}{2}\).

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Most popular questions from this chapter

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