Math Problem Statement

Considere la siguiente suma: \( \sum_{n=2}^{\infty} 7 a^n = 10 \). El valor positivo de \( a \) para que se cumpla la ecuación es:

Solution

The problem asks us to find the positive value of aa such that the following equation is satisfied:

n=27an=10\sum_{n=2}^{\infty} 7 a^n = 10

This is an infinite geometric series with the general form:

n=2ran=rn=2an\sum_{n=2}^{\infty} r a^n = r \sum_{n=2}^{\infty} a^n

We can rewrite the series as:

7a2+7a3+7a4+=107 a^2 + 7 a^3 + 7 a^4 + \dots = 10

Factoring out 7, we get:

7(a2+a3+a4+)=107 (a^2 + a^3 + a^4 + \dots) = 10

The expression inside the parentheses, a2+a3+a4+a^2 + a^3 + a^4 + \dots, is itself an infinite geometric series with the first term a2a^2 and common ratio aa. The sum of an infinite geometric series with first term AA and common ratio rr (where r<1|r| < 1) is given by:

A1r\frac{A}{1 - r}

In this case, A=a2A = a^2 and r=ar = a, so we can write:

a2+a3+a4+=a21aa^2 + a^3 + a^4 + \dots = \frac{a^2}{1 - a}

Substituting back, we have:

7a21a=107 \cdot \frac{a^2}{1 - a} = 10

Now, solving for aa:

  1. Divide both sides by 7: a21a=107\frac{a^2}{1 - a} = \frac{10}{7}

  2. Cross-multiply to eliminate the fraction: a2=107(1a)a^2 = \frac{10}{7} (1 - a)

  3. Distribute 107\frac{10}{7}: a2=107107aa^2 = \frac{10}{7} - \frac{10}{7}a

  4. Rearrange the terms to form a standard quadratic equation: a2+107a107=0a^2 + \frac{10}{7}a - \frac{10}{7} = 0

  5. Multiply through by 7 to clear the fractions: 7a2+10a10=07a^2 + 10a - 10 = 0

Now, we can solve this quadratic equation using the quadratic formula:

a=b±b24ac2aa = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

where a=7a = 7, b=10b = 10, and c=10c = -10.

Plugging in these values:

a=10±10247(10)27a = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 7 \cdot (-10)}}{2 \cdot 7}

a=10±100+28014a = \frac{-10 \pm \sqrt{100 + 280}}{14}

a=10±38014a = \frac{-10 \pm \sqrt{380}}{14}

a=10±29514a = \frac{-10 \pm 2\sqrt{95}}{14}

Simplifying further, we get:

a=5±957a = \frac{-5 \pm \sqrt{95}}{7}

Since we are asked for the positive value of aa, we take the positive root:

a=5+957a = \frac{-5 + \sqrt{95}}{7}

Let me know if you would like further clarification on any steps or a numeric approximation for this solution.


Here are 5 related questions for further exploration:

  1. How do you find the sum of an infinite geometric series starting from a general term?
  2. What is the convergence condition for an infinite geometric series?
  3. Can the quadratic formula be used to solve all types of quadratic equations?
  4. How does changing the common ratio affect the sum of an infinite series?
  5. Why is only the positive root considered in this context?

Tip: When working with infinite series, always verify if the common ratio's absolute value is less than 1 to ensure convergence.

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Math Problem Analysis

Mathematical Concepts

Infinite Geometric Series
Algebra
Quadratic Equations

Formulas

Sum of an infinite geometric series \( \frac{A}{1 - r} \)
Quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Theorems

Convergence of infinite geometric series
Quadratic formula theorem

Suitable Grade Level

Grades 10-12