Consider a geometric sequence with the first term a and common ratio |r| < 1. Then the n-th partial sum (the sum of the first n terms) of the sequence is

$S_n=a+ar+ar^2+\cdots+ar^{n-2}+ar^{n-1}$

Multiply both sides by r :

$rS_n=ar+ar^2+ar^3+\cdots+ar^{n-1}+ar^n$

Subtract the latter sum from the first, which eliminates all but the first and last terms:

$S_n-rS_n=a-ar^n$

Solve for $S_n$ :

$(1-r)S_n=a(1-r^n)\implies S_n=\dfrac a{1-r}-\dfrac{ar^n}{1-r}$

Then as gets arbitrarily large, the term $r^n$ will converge to 0, leaving us with

$S=\displaystyle\lim_{n\to\infty}S_n=\frac a{1-r}$

So the given series converge to

(I) -243/(1 + 1/9) = -2187/10

(II) -1.1/(1 + 1/10) = -1

(III) 27/(1 + 1/3) = 18

8 0

2 years ago

You're given two side lengths of 6 centimeters and 9 centimeters. Which measurement can you use for the length of the third side

Zarrin [17]

Does it have to be a different number if not do nine cause then the bottom line soul be 6 and the sides would be 9

7 0

3 years ago