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Find the sum of the first 38 terms of the following series, to the nearest integer.

3,9,15,dots
Answer:

Find the sum of the first $38$ terms of the following series, to the nearest integer. $\newline$ $3,9,15, \ldots$ $\newline$ Answer:

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Geometric sequences

Full solution

Q. Find the sum of the first

38

terms of the following series, to the nearest integer.

\newline

3,9,15, \ldots

\newline

Answer:

Identify type and difference: Identify the type of series and the common difference. $\newline$ The given series is arithmetic because there is a constant difference between consecutive terms. $\newline$ To find the common difference $d$ , subtract the first term from the second term. $\newline$ $d = 9 - 3 = 6$
Use arithmetic series formula: Use the formula for the sum of an arithmetic series. $\newline$ The sum of the first $n$ terms of an arithmetic series can be found using the formula: $\newline$ $S_n = \frac{n}{2} \times (2a + (n - 1)d)$ $\newline$ where $S_n$ is the sum of the first $n$ terms, $a$ is the first term, and $d$ is the common difference.
Plug in values: Plug in the values into the formula to find the sum of the first $38$ terms. $\newline$ Let's use the formula with $n = 38$ , $a = 3$ , and $d = 6$ . $\newline$ $S_{38} = \frac{38}{2} \times (2 \times 3 + (38 - 1) \times 6)$
Simplify expression: Simplify the expression to find the sum. $\newline$ $S_{38} = 19 \times (6 + 37\times6)$ $\newline$ $S_{38} = 19 \times (6 + 222)$ $\newline$ $S_{38} = 19 \times 228$ $\newline$ $S_{38} = 4332$

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