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Find the sum of the first 48 terms of the following series, to the nearest integer.

2,7,12,dots
Answer:

Find the sum of the first $48$ terms of the following series, to the nearest integer. $\newline$ $2,7,12, \ldots$ $\newline$ Answer:

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Solve exponential equations using logarithms

Full solution

Q. Find the sum of the first

48

terms of the following series, to the nearest integer.

\newline

2,7,12, \ldots

\newline

Answer:

Identify Terms and Numbers: Identify the first term $a_1$ , common difference $d$ , and number of terms $n$ in the arithmetic series. $\newline$ The first term $a_1$ is $2$ , the common difference $d$ is $7 - 2 = 5$ , and the number of terms $n$ is $48$ .
Use Arithmetic Series Formula: Use the formula for the sum of the first $n$ terms of an arithmetic series: $S_n = \frac{n}{2} \times (2a_1 + (n - 1)d)$ . We will plug in the values for $a_1$ , $d$ , and $n$ into the formula.
Calculate Expression Inside Parentheses: Calculate the sum using the values: $S_{48} = \frac{48}{2} \times (2\times2 + (48 - 1)\times5)$ . First, calculate the expression inside the parentheses: $2\times2 + 47\times5 = 4 + 235 = 239$ .
Calculate Sum: Now, calculate the sum: $S_{48} = 24 \times 239$ . Perform the multiplication: $S_{48} = 5736$ .
Round to Nearest Integer: Round the sum to the nearest integer. The sum $S_{48}$ is already an integer, so rounding is not necessary.

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