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

Question

Lily Anderson · Accepted Answer

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!

Elias Ackerman · Answer

undefined 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.