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Find the sum of the first 10 terms of the following series, to the nearest integer.

6,12,24,dots
Answer:

Find the sum of the first $10$ terms of the following series, to the nearest integer. $\newline$ $6,12,24, \ldots$ $\newline$ Answer:

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Geometric sequences

Full solution

Q. Find the sum of the first

10

terms of the following series, to the nearest integer.

\newline

6,12,24, \ldots

\newline

Answer:

Identify type and ratio: Identify the type of series and the common ratio. $\newline$ The series is geometric because each term is a multiple of the previous term. To find the common ratio, divide the second term by the first term. $\newline$ $\frac{12}{6} = 2$ $\newline$ Common Ratio: $2$
Use sum formula: Use the formula for the sum of the first $n$ terms of a geometric series. $\newline$ The formula for the sum of the first $n$ terms ( $S_n$ ) of a geometric series is: $\newline$ $S_n = \frac{a(1 - r^n)}{(1 - r)}$ , where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.
Plug values into formula: Plug the values into the formula to find the sum of the first $10$ terms. $a = 6$ (the first term) $r = 2$ (the common ratio) $n = 10$ (the number of terms) $S_{10} = \frac{6(1 - 2^{10})}{(1 - 2)}$
Calculate sum: Calculate the sum using the values from Step $3$ . $\newline$ $S_{10} = 6(1 - 1024) / (1 - 2)$ $\newline$ $S_{10} = 6(-1023) / (-1)$ $\newline$ $S_{10} = 6 \times 1023$ $\newline$ $S_{10} = 6138$

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