Question

find the sum of the first 88 terms of the following sequence. round to the nearest hundredth if necessary. note: $s_n = a_1\frac{1 - r^n}{1 - r}$ 45,15,5,... \bigcirc 69.00 \bigcirc 135.00 \bigcirc 67.50 \bigcirc 66.75

Explanation:

Step1: Identify the sequence type

The sequence is \(45, 15, 5, \dots\). It's a geometric sequence because each term is multiplied by a common ratio \(r\). To find \(r\), divide the second term by the first term: \(r=\frac{15}{45}=\frac{1}{3}\), and check with the third term: \(\frac{5}{15}=\frac{1}{3}\), so \(r = \frac{1}{3}\), \(a_1 = 45\), and \(n = 88\).

Step2: Use the geometric series sum formula

The formula for the sum of the first \(n\) terms of a geometric series is \(S_n=a_1\frac{1 - r^n}{1 - r}\) (for \(|r| < 1\), as \(n\) is large, \(r^n\) approaches 0). Here, \(a_1 = 45\), \(r=\frac{1}{3}\), \(n = 88\). Since \(|r|=\frac{1}{3}<1\), as \(n\) becomes very large (like 88), \(r^{88}\) is extremely close to 0. So we can approximate the sum as \(S_{\infty}=\frac{a_1}{1 - r}\) (the sum of an infinite geometric series, which is a good approximation here because \(r^{88}\) is negligible).

Step3: Calculate the sum

Substitute \(a_1 = 45\) and \(r=\frac{1}{3}\) into the formula:

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Answer:

67.50