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1) Find the sum of the first 50 terms 1, 8, 15,... using the sum of an arithmetic series formula.

2) Find the sum of the n terms of the arithmetic sequence a1 = 7, a12 = 29, n = 12.

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1) Find the sum of the first 50 terms 1, 8, 15,... using the sum of an arithmetic series formula.

2) Find the sum of the n terms of the arithmetic sequence a1 = 7, a12 = 29, n = 12.

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chroot

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symbolipoint

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Learn the derivations of the formulas if you have trouble remembering the formulas. This could help you to remember the formulas. For arithmetic series & sequences, even just remembering the formulas should not be very difficult. Study more; do more exercises.

1) Find the sum of the first 50 terms 1, 8, 15,... using the sum of an arithmetic series formula.

2) Find the sum of the n terms of the arithmetic sequence a1 = 7, a12 = 29, n = 12.

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symbolipoint

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Buy a used book; they are cheap. No more than $2 at a public library used book depository can yield any of a good set of intermediate or college algebra textbooks. These books are very easy to find in used supplies. If the book is 20 or 30 or more years old, the concepts and skills will be the same; so will the formulas, all of which will usually be derived for you in the book.

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You wouldn't happen to have any examples of arithmetic series?

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Just any example?You wouldn't happen to have any examples of arithmetic series?

Like this:

1+2+3+4+5+6+7+8+9+10=55=(10)(10+1)/2

In general an arithmetic series arises by adding together the terms in an arithmetic sequence.

In an arithmetic sequence we start with an initial term let's call it a, and then to get the next term we add a number (I think usually called the common difference, I'm not positive though) lets call it d.

So we have a sequence of numbers a

So that

a

And

a

To clarify why d is usually called a common difference consider the difference

a

Then from this sequence we get an arithmetic series by adding up all of these terms, so the series is a

It would be a good excercise to derive the sum of a general arithmetic series from the sum of the first n natural numbers:

1+2+3+...+n=n(n+1)/2

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