Week 1 – Math 11 – Arithmetic sequences

to find a value in your sequence (t_{n}) you must use this formula…

t_{n}t_{1} + (n – 1)(d)

to give an example here is my pattern – 3, 7, 11,  15, 19…
to find t_{50} follow these steps….

the first thing you will need in your formula is t_{1}
in this sequence 3 is t_{1} because it is the first number of the sequence, so your formula so far would look like this…
t_{50} = 3 + (n – 1)(d)

next, you will want to subtract 1 from 50, because the next part of our formula is (n – 1). the (n) represents the number in your sequence just as the second number of your sequence is represented by  t_{2}. so your formula should look like this after you subtracted 1 from 50…
t_{50} = 3 + (49)(d)

the final thing in your sequence is (d). the (d) represents the common difference, the common difference is the rate increasing between numbers. in our sequence, our common difference is 4 because we increase 4 everytime. ex/ 3 +4 = 7 +4 = 11 +4 = 15.
once you know your common difference your formula should look like this…
t_{50} = 3 + (49)(4)

once you calculate your formula it should look something like this…

t_{50} = 3 + (49)(4) = 199

Part 2
to find the arithmetic series you need to use this formula…

S_{n}\frac{n}{2}(t_{1}t_{n})

first, we will start by adding what we know this far…
S_{50}\frac{50}{2}(3 + t_{n})

The last thing you need is t_{n}. since we are in S_{50} we need to use the awnser of t_{50}. Once you add this to your formula it should look like this…
S_{50}\frac{50}{2}(3 + 199)

When you add this all up  you should get this…

S_{50}\frac{50}{2}(3 + 199) = 5,050

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