Category Archives: Math 11

Week 2 – Math 11 – Geometric series

To find a term in a geometric sequence you must use the formula

t_{n} = a (r)^{(n-1)}

to give an example here is my pattern  -3, 9, -27, 81…
to find  t_{10} follow these steps….

First, (r)^{(n-1)} becomes (r)^{(10-1)} wich is (r)^{9} leaving you with…

t_{10} = a (r)^{9}

Then you want to add in (a), replace (a) with -3 because it’s your first term. making your equation look like…

t_{10} = -3 (r)^{9}

once you have used all the given numbers you have. find the common ratio (r). if you follow the pattern of the sequence you will notice it is being multiplied by (-3) each time. so replace (r) with (-3)

t_{10} = -3 (-3)^{9}

(-3)^{9} is -19683

-3 x -19683 is 59,049

therefore t_{10} = 59,049

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

S_{n} = \frac{a (r^{n}-1)}{r-1}

using the pattern above we will find the series of  S_{10}. first lets add in the values we were given straight of the bat. replace (a) with -3, and replace (n) with 10…

S_{10} = \frac{-3 (r^{10}-1)}{r-1}

if you remember, above we found out that the common ratio was -3, so replace (r) with -3…

S_{10} = \frac{-3 (-3^{10}-1)}{-3-1}

first, we will do(-3)^{10} wich is,  59,049. then subtract 1 from this giving you, 59,048. now add this in…

S_{10} = \frac{-3 (59,048)}{-3-1}

after that multiply 59,048 by -3 giving you, -177,144. now add this to your equation…

S_{10} = \frac{-177,144}{-3-1}

now we must solve the denominator, -3-1 = -4.  your equation should look like this…

S_{10} = \frac{-177,144}{-4}

now divide these…

S_{10} = 44,286

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

Week1 – Precal 11 Arithmetic sequence

Arithmetic sequences add the same amount to each time. ex/ 7, 10, 13, 16…

Common difference stands for the amount increasing and is represented by (d). ex/ 7,10,13 d=3

Tn stands for the unknown number you’re trying to find, Tn could be the last number in the sequence or the 50th number.

T is used to represent the numbers in the sequence. ex/
7, 10, 13, 16…
T1,T2,T3,T4