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Find the sum of the first 43 terms of the following series, to the nearest integer.

4,12,20,dots
Answer:

Find the sum of the first $43$ terms of the following series, to the nearest integer. $\newline$ $4,12,20, \ldots$ $\newline$ Answer:

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Geometric sequences

Full solution

Q. Find the sum of the first

43

terms of the following series, to the nearest integer.

\newline

4,12,20, \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, subtract the first term from the second term: $\newline$ $12 - 4 = 8$ $\newline$ Common Difference: $8$
Find $43$ rd Term: Find the $43^{\text{rd}}$ term of the series using the formula for the nth term of an arithmetic sequence: $a_n = a_1 + (n - 1)d$ , where $a_n$ is the nth term, $a_1$ is the first term, $n$ is the term number, and $d$ is the common difference. $\newline$ $a_{43} = 4 + (43 - 1) \times 8$ $\newline$ $a_{43} = 4 + 42 \times 8$ $\newline$ $a_{43} = 4 + 336$ $\newline$ $a_{43} = 340$
Calculate Sum of $43$ Terms: Calculate the sum of the first $43$ terms using the formula for the sum of an arithmetic series: $S_n = \frac{n}{2} * (a_1 + a_n)$ , where $S_n$ is the sum of the first $n$ terms. $\newline$ $S_{43} = \frac{43}{2} * (4 + 340)$ $\newline$ $S_{43} = 21.5 * 344$ $\newline$ $S_{43} = 7396$

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