A geometric progression has first term a (where a>0), and common ratio r. The sum of the first n terms is?

Question

Part 1: A geometric progression has first term a (where a>0), and common ratio r. The sum of the first n terms is S(n) and the sum to infinity is S. Given that S(2) is twice the valus of the fifth term, find the value of r. Hence, find the least value of n such that S(n) is within 5% of S.

Part 2: It is given that the solution set of 1/(1+x)

How can r=1 when there is a sum to infinity? Isn't IrI<1?

Answer 1

Part 2 is easy. The function y = 1/(1+x) has an infinite discontinuity at x = -1. Also, y(0) = 1, and for large x (positive and negative), the function approaches the x-axis as an asymptote. For positive x, y(x) > 0, and for large negative x, y(x) < 0. It's easy to draw the graph.

In this problem, we care about y(-3) = -1/2; y(-2) = -1; and y(1) = 1/2. We want the parabola y = ax^2 + bx + c that passes through the points (-3, -1/2), (-2, -1), and (1, 1/2). Since all three points must satisfy the quadratic equation, we can plug in values and write these three equations:

(1) For x = -3: 9a - 3b + c = -1/2
(2) For x = -2: 4a - 2b + c = -1
(3) For x = 1: a + b + c = 1/2

The solution to this set of three linear equations in three unknowns is a = 1/4; b = 3/4; and c = -1/2. (Details of the solution omitted.)

The quadratic is y = 1/4 x^2 + 3/4 x - 1/2. This function passes through the three points we identified, and

1/(1+x) < 1/4 x^2 + 3/4 x - 1/2 over the range specified in the problem. (Actually, there's equality at three points.)

That inequality can also be written as

4/(1+x) < x^2 + 3x - 2

I'll answer the other part of the question as best I can in the morning. Right now, it's time to sleep.

Answer 2

Sn = a + ar + ar ² + -------------ar^(n-a million) rSn = __a r + ar ² + ---------------------------ar^n ( a million - r ) Sn = a ( a million - r^n ) Sn = a ( a million - r^n ) / ( a million - r ) OR Sn = a ( r^n - a million ) / ( r - a million )