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Find the sum of the first 41 terms of the following series, to the nearest integer.

8,14,20,dots
Answer:

Find the sum of the first $41$ terms of the following series, to the nearest integer. $\newline$ $8,14,20, \ldots$ $\newline$ Answer:

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Geometric sequences

Full solution

Q. Find the sum of the first

41

terms of the following series, to the nearest integer.

\newline

8,14,20, \ldots

\newline

Answer:

Identify type of series: Identify the type of series. $\newline$ The series $8, 14, 20, \ldots$ is an arithmetic series because the difference between consecutive terms is constant.
Determine common difference: Determine the common difference of the series. $\newline$ The difference between the second term $14$ and the first term $8$ is $14 - 8 = 6$ . $\newline$ Common Difference $d$ : $6$
Find first term: Find the first term of the series. $\newline$ The first term $a_1$ of the series is given as $8$ . $\newline$ First Term $a_1$ : $8$
Use formula for sum: Use the formula for the sum of the first $n$ terms of an arithmetic series. $\newline$ The sum of the first $n$ terms ( $S_n$ ) of an arithmetic series is given by the formula: $\newline$ $S_n = \frac{n}{2} \times (2a_1 + (n - 1)d)$ $\newline$ Where: $\newline$ $S_n$ = sum of the first $n$ terms $\newline$ $n$ = number of terms $\newline$ $a_1$ = first term $\newline$ $d$ = common difference
Calculate sum of terms: Calculate the sum of the first $41$ terms using the formula. $\newline$ Substitute $n = 41$ , $a_1 = 8$ , and $d = 6$ into the formula: $\newline$ $S_{41} = \frac{41}{2} * (2*8 + (41 - 1)*6)$ $\newline$ $S_{41} = 20.5 * (16 + 40*6)$ $\newline$ $S_{41} = 20.5 * (16 + 240)$ $\newline$ $S_{41} = 20.5 * 256$ $\newline$ $S_{41} = 5252$
Round sum: Round the sum to the nearest integer. $\newline$ The sum of the first $41$ terms is $5252$ , which is already an integer, so no rounding is necessary.

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