# SOLUTION: Use the geometric sequence of numbers 1, 1/3, 1/9 , 1/27… to find the following: a) What is r, the ratio between 2 consecutive terms? Answer: Show work in this space.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Use the geometric sequence of numbers 1, 1/3, 1/9 , 1/27… to find the following: a) What is r, the ratio between 2 consecutive terms? Answer: Show work in this space.       Log On

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 Quadratics: solvers Practice! Answers archive Lessons Word Problems In Depth

 Question 87238: Use the geometric sequence of numbers 1, 1/3, 1/9 , 1/27… to find the following: a) What is r, the ratio between 2 consecutive terms? Answer: Show work in this space. b) Using the formula for the sum of the first n terms of a geometric sequence, what is the sum of the first 10 terms? Carry all calculations to 6 decimals on all assignments. Answer: Show work in this space. c) Using the formula for the sum of the first n terms of a geometric sequence, what is the sum of the first 12 terms? Carry all calculations to 6 decimals on all assignments. Answer: Show work in this space. d) What observation can make about the successive partial sums of this sequence? In particular, what number does it appear that the sum will always be smaller than? Answer: Answer by jim_thompson5910(29613)   (Show Source): You can put this solution on YOUR website!a) The ratio r is the factor to get from term to term. So to find r, simply pick any term and divide it by the previous term: Divide the term of by So the ratio is The sequence is reduced by a factor of each term, so the sequence is b) The sum of a geometric series is where a=1 So plug in and n=10 to find the sum of the first 10 partial sums Raise to the 10 power Make "1" into an equivalent fraction with a denominator of 59049 Subtract the fractions in the numerator Subtract the fractions in the denominator Flip the 2nd fraction, multiply, and simplify So the sum of the first ten terms is or 1.49997459736829 approximately note: I chose to use fractions (to maintain accuracy), but it may be much easier for you to simply use a calculator to evaluate the sum. c) The sum of a geometric series is where a=1 So plug in and n=12 to find the sum of the first 12 partial sums Raise to the 10 power Make "1" into an equivalent fraction with a denominator of 531441 Subtract the fractions in the numerator Subtract the fractions in the denominator Flip the 2nd fraction, multiply, and simplify So the sum of the first twelve terms is or 1.49999717748537 approximately d) It appears that the sums are approaching a finite number of or 1.5. This is because each term is getting smaller and smaller. This observation is justified by the fact that if then the infinite series will approach a finite number. In other words If (the magnitude of r has to be less than 1) then, Where S is the sum of the infinite series. So if we let a=1 and r=1/3 we get So this verifies that our series approaches .