How do you find the sum of the first 20 terms of the series #3+8+13+18+23+...#?

Answer 1

Lily Anderson

The first twenty terms have a sum of #1010#.

In this problem, #a = 3#, #r = 8 - 3 = 5#, #n = 20# and #s_20 = ?#.

We will use the formula #s_n = n/2(2a + (n - 1)d)#

#s_20 = 20/2(2(3) + (20 - 1)5)#

#s_20 = 10(6 + 95)#

#s_20 = 10(101)#

#s_20 = 1010#

Hopefully this helps!

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Answer 2

Elias Ackerman

To find the sum of the first 20 terms of the series $3+8+13+18+23+...$ , use the formula for the sum of an arithmetic series:

$S = \frac{n}{2}(a_1 + a_n)$

Where:

$S$ is the sum of the series,
$n$ is the number of terms in the series (which is 20 in this case),
$a_1$ is the first term of the series, and
$a_n$ is the last term of the series.

In this series:

$a_1 = 3$ (the first term),
To find $a_n$ , we use the formula for the $n$ th term of an arithmetic sequence: $a_n = a_1 + (n - 1)d$ , where $d$ is the common difference between consecutive terms.
Since the common difference between consecutive terms is 5 (each term increases by 5), we have $d = 5$ .
Now, plug in $n = 20$ into the formula for $a_n$ : $a_{20} = 3 + (20 - 1) \times 5 = 3 + 19 \times 5 = 3 + 95 = 98$

Using the formula for the sum of an arithmetic series: $S = \frac{n}{2}(a_1 + a_n)$ $S = \frac{20}{2}(3 + 98)$ $S = \frac{20}{2}(101)$ $S = 10 \times 101$ $S = 1010$

So, the sum of the first 20 terms of the series $3+8+13+18+23+...$ is 1010.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.