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Find the sum of the following arithmetic series: 3 + 8 + 13 + 18 + . . . + 48

Find the sum of the following arithmetic series: 3 + 8 + 13 + 18 + . . . + 48-example-1
User Ctc Chen
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1 Answer

20 votes
20 votes

Answer:

The sum of this series is 255

Step-by-step explanation:

First, we need to identify the equation for the arithmetic series, so it has the form

an = a1 + (n-1)d

Where an is the nth term, a1 is the first term and d is a common difference.

In this case, a1 = 3 and the common difference is 5 because

8 - 3 = 5

13 - 8 = 5

18 - 13 = 5

Then, the equation for the arithmetic series is

an = 3 + (n-1)5

Now, let's identify the position of the term 48, so replacing an = 48 and solving for n, we get:

48 = 3 + (n - 1)5

48 - 3 = (n - 1)5

45 = (n - 1)5

45/5 = n - 1

9 = n - 1

9 + 1 = n

10 = n

Therefore, 48 is the 10th term of the series.

Finally, the sum of the first n terms of an arithmetic series is equal to

Sn = n(a1 + an)/2

So, replacing n = 10, a1 = 3, and an = 48, we get that the sum of this series is

Sn = 10(3 + 48)/2

Sn = 10(51)/2

Sn = 510/2

Sn = 255

So, the answer is 255

User Argentpepper
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