What is the sum of the geometric sequence 8, 16, 32?

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Answer 1 · 2014-07-28T17:06:20

The sum of the geometric sequence is 56.

To find the sum of any geometric sequence, you use the equation: $S_{n} = \frac{a (r^{n} - 1)}{r - 1}$ $S_n = (a(r^n -1))/(r-1)$ where:

a --> is the first term of the sequence; in this case "a" is 8.

r --> is the ratio (what each number is being multiplied by) between each number in the sequence; in this case, each term is being multiplied by 2.

n --> is the number of terms in the sequence; in this case, there are 3 terms.

Knowing the value of each variable, you can substitute the values into the equation as follows:

$S_{n} = \frac{8 (2^{3} - 1)}{2 - 1}$ $S_n = (8(2^3 -1))/(2-1)$

Simplify and solve:

$S_{n} = \frac{56}{1}$ $S_n = (56)/(1)$

Therefore, the sum of the sequence is 56! Hopefully you've understood this and hopefully I was of some help! :)