Question

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Answer 1

The function A(n) = 22(1.1)^n-1 is an illustration of a geometric series

The sum of the 23rd through 40th terms of the series is 17.49

How to determine the sum of the series

The nth term of the geometric series is given as:

$A(n) = 22(1.1)^{n-1$

The nth term of a series is represented as:

$A(n) = ar^{n-1$

So, by comparison;

We have:

$a = 22$

$r = 1.1$

The sum of nth term of a geometric progression is:

$S_n = \frac{a(r^n - 1)}{r - 1}$

Start by calculating the sum of the first 22 terms

$S_{22} = \frac{22(1.1^{22} - 1)}{22 - 1}$

$S_{22} = \frac{157.086}{21}$

$S_{22} = 7.48$

Next, calculate the sum of the first 40 terms

$S_{40} = \frac{22(1.1^{40} - 1)}{40 - 1}$

$S_{40} = \frac{973.70}{39}$

$S_{40} = 24.97$

Subtract S22 from S40

$S_{40} - S_{22} = 24.97 - 7.48$

$S_{40} - S_{22} = 17.49$

Hence, the sum of the 23rd through 40th terms of the series is 17.49

Read more about progression at:

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