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in this video we're going to focus
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mostly on arithmetic sequences
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now to understand what an arithmetic
00:00:07
sequence is it's helpful to distinguish
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it from a geometric sequence
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so here's an example of an arithmetic
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sequence
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the numbers 3 7 11
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15
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19
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23 and 27 represents an arithmetic
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sequence
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this would be a geometric sequence 3
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6
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12
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24
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48
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96 192.
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do you see the difference between these
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two sequences and do you see any
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patterns within them
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in the arithmetic sequence on the left
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notice that we have a common difference
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this is the first term this is the
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second term this is the third fourth and
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fifth term
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to go from the first term to the second
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term we need to add four
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to go from the second to the third term
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we need to add four
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and that is known as the common
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difference
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in a geometric sequence you don't have a
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common difference rather you have
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something that is called the common
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ratio to go from the first term to the
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second term
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you need to multiply by two
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to go from the second term to the third
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term you need to multiply by two again
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so that is the r value that is the
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common ratio
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so in an arithmetic sequence the pattern
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is based on addition and subtraction
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in a geometric sequence the pattern is
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based on multiplication and
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division now the next thing that we need
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to talk about
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is
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the mean
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how to calculate the arithmetic mean and
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the geometric mean
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the arithmetic mean is basically the
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average of two numbers
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it's a plus b divided by two
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so when taking
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an arithmetic mean of two numbers within
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an arithmetic sequence let's say if we
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were to take
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the mean of three and eleven
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we would get the middle number in that
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sequence in this case we would get 7.
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so if you were to add 3 plus 11
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and divide by 2 3 plus 11 is 14 14
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divided by 2 gives you 7.
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now let's say if we wanted to find the
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arithmetic mean between 7
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and 23 it's going to give us the middle
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number
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of that sequence which is 15.
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so if you would add up 7
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plus 23 divided by 2
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7 plus 23 is 30 30 divided by 2 is 15.
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so that's how you can calculate the
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arithmetic mean and that's how you can
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identify it within
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an arithmetic sequence
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the geometric mean
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is the square root
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of a times b
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so let's say if we want to find the
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geometric mean between three and six
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it's going to give us the middle number
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of the sequence which is
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i mean if we were to find the geometric
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mean between 3 and 12
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we will get the middle number of that
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sequence which is 6.
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so in this case a is 3 b is 12. 3 times
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12 is 36
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the square root of 36 is 6.
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now let's try another example
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let's find the geometric mean
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between 6 and 96
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this should give us the middle number
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24.
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now we need to simplify this radical
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96 is six times sixteen
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six times six is thirty six
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the square root of thirty six is six the
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square root of sixteen is four
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so we have six times four
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which is twenty 24.
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so as you can see
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the geometric mean of two numbers within
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the geometric sequence will give us the
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middle number in between those two
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numbers in that sequence
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now let's clear away a few things
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the formula that we need to find the nth
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term
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of an arithmetic sequence is a sub n
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is equal to a sub 1
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plus n minus 1 times the common
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difference d
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in a geometric sequence
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its a sub n
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is equal to a 1
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times r
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raised to the n minus 1.
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now
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let's use that equation to get the fifth
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term in the arithmetic sequence
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so that's going to be a sub 5
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a sub 1 is the first term which is three
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n is five since we're looking for the
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fifth term the common difference
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is four in this problem
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five minus one is four
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4 times 4 is 16 3 plus 16 is 19.
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so this formula
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gives you any term in the sequence you
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could find the fifth term the seventh
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term the 100th term and so forth
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now in a geometric sequence
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we could use this formula
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so let's calculate the six term
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of the geometric sequence it's going to
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be a sub six
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which equals a sub one the first term is
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three the common ratio is two
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and this is going to be raised to the
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six minus one
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six minus one is five
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and then two to the fifth power if you
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multiply two five times two times two
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times two times two times two
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so we can write it out
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so this here that's four
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three twos make eight four times eight
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is thirty two
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so this is three
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times thirty two
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three times thirty is ninety three times
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two is six
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so this will give you
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ninety six
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so that's how you could find the f term
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in a geometric sequence
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by the way
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make sure you have a sheet of paper to
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write down these formulas
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so that when we work on some practice
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problems
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you know what to do
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now the next thing we need to do is
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be able to calculate the partial sum of
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a sequence
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s sub n is the partial sum of
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a series of a few terms
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and it's equal to the first term plus
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the last term
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divided by two times n
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for geometric sequence the partial sum s
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of n is going to be a sub 1
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times 1 minus r raised to the n
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over
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1 minus r
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so let's find the sum
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of the first seven terms in this
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sequence
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so that's going to be s sub 7
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that's going to equal the first term
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plus the seventh term
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divided by 2
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times n
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where n is the number of terms which is
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7.
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now think about what this means
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so basically to find the sum of an
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arithmetic sequence
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you're basically taking the average
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of the first and the last term in that
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sequence and then multiplying it by the
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number of terms
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in that sequence
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because this is basically the average of
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3
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and 27.
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and we know the average or the
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arithmetic mean of 3 and 27 that's going
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to be the middle number 15.
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so let's go ahead and plug this in
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so this is 3 plus 27
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over 2
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times 7.
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3 plus 27 is 30 plus 2 i mean well 30
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divided by 2 that's 15.
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so the average
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of the first and last term is 15 times 7
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10 times 7 is 70. 5 times 7 is 35
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so this is going to be 105.
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that's the sum of the first seven terms
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and you can confirm this with your
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calculator if you add up three plus
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seven
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plus eleven
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plus fifteen
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plus 19 plus 23
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and then plus 27
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and that will give you s of 7 the sum of
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the first seven terms
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go ahead and add up those numbers
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if you do you'll get 105.
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so that's how you can confirm your
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answer
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now let's do the same thing with a
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geometric sequence
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so let's get the sum
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of the first six terms
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s sub six
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so this is going to be three
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plus six
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plus twelve
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plus twenty four
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plus 48
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plus 96
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so we're adding the first six terms
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now because it's not many terms we're
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adding we can just simply plug this into
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our calculator
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and we'll get 189
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but now let's confirm this answer using
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the formula
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so s sub 6 the sum of the first six
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terms
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is equal to the first term a sub one
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which is three
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times one minus r r is the common ratio
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which is two
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raised to the n
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n is six
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over one minus r or 1 minus 2.
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so i'm going to work over here since
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there's more space
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now 2 to the 6
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that's going to be 64. if you recall 2
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to the fifth power was 32 if you
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multiply 32 by 2 you get 64.
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so this is going to be 1 minus 64.
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and 1 minus 2 is negative 1.
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so this is 3 times
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1 minus 64.
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is negative 63.
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so we could cancel the two negative
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signs a negative divided by a negative
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will be a positive
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so this is just 3 times 63.
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3 times 6 is 18. so 3 times 60 has to be
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180
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and then 3 times 3 is 9 180 plus 9
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adds up to 189
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so we get the same answer
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now what is the difference between
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a sequence and a series
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i'm sure you heard of these two terms
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before but what is the difference
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between them
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now we've already considered what
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an arithmetic sequence is
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a sequence is basically a list of
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numbers
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so that's a sequence
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a series is the sum
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of the numbers in a sequence
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so this here is
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an arithmetic sequence
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this
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is an arithmetic series because it's the
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sum of an arithmetic sequence
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now what we have here is a sequence but
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it's a geometric sequence as we've
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considered earlier
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this
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is a geometric series it's the sum of a
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geometric sequence
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now there are two types of sequences and
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two types of series
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you have a finite sequence
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and an infinite sequence and it's also a
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finite series and an infinite series
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this sequence is finite it has a
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beginning and it has an n
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this series is also finite it has a
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beginning and it has an end
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in contrast if i were to write 3 7
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11
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15 19 and then dot dot dot
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this would be
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an infinite sequence
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the presence of these dots tells us that
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the numbers keep on going to infinity
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now the same is true for a series
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let's say if i had
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three plus seven
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plus eleven plus fifteen plus nineteen
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and then plus dot dot dot dot
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that would also be
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an infinite series
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so now you know the difference between a
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finite series and an infinite series
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now let's work on some practice problems
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describe the pattern of numbers shown
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below is it a sequence or series
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is it finite or infinite
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is it
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arithmetic geometric or neither
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so let's focus on if it's a sequence or
00:14:14
series first
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part a
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so we got the numbers 4 7 10 13 16 19.
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we're not adding the numbers we're
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simply making a list of it so this is
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a
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sequence the same is true for part b
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we're simply listing the numbers so
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that's a sequence
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in part c we're adding a list of numbers
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so since we have a sum this is going to
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be a series
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d is also a series
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e
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that's a sequence
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for f we're adding numbers so that's a
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series and the same is true
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for g
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so hopefully this example helps you to
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see the difference between a sequence
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and a series
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now let's move on to the next topic is
00:15:04
it finite or is it infinite
00:15:10
to answer that all we need to
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do is identify if we have a list of dots
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at the end or not
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here this ends at 19. so that's a finite
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sequence
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the dots here tells us it's going to go
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forever so this is an infinite
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sequence
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this one we have the dots so this is
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going to be an infinite series
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this ends at 162 so it's finite so we
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have a finite series
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this is going to be an infinite sequence
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next we have an infinite series
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and the last one is a finite series
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now let's determine if we're dealing
00:16:02
with
00:16:03
an arithmetic geometric
00:16:06
or neither sequence or series
00:16:09
so we're looking for a common difference
00:16:11
or a common ratio
00:16:14
so for a
00:16:15
notice that we have a common difference
00:16:17
of three four plus three is seven
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seven plus three is ten
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so because we have a common difference
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this is going to be
00:16:26
an arithmetic sequence
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for b going from the first number to the
00:16:34
second number we need to multiply by two
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four times two is eight eight times two
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is sixteen
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so we have a common ratio
00:16:43
which makes this sequence geometric
00:16:50
for answer choice c
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going from five to nine that's plus four
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and from nine to thirteen that's plus
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four so we have a common difference
00:17:00
so this is going to be not an arithmetic
00:17:02
sequence
00:17:03
but an arithmetic series
00:17:08
for answer choice d going from two to
00:17:10
six we're multiplying by three and then
00:17:12
six times three is eighteen
00:17:15
so that's a geometric
00:17:18
a geometric series
00:17:22
now for e
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going from 50 to 46 that's a difference
00:17:26
of negative 4 and 46 to 42 that's the
00:17:30
difference of negative four so this
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is arithmetic
00:17:39
for f
00:17:40
we have a common ratio of four three
00:17:42
times four is twelve
00:17:44
twelve times 4 is 48
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and if you're wondering how to calculate
00:17:49
d and r
00:17:50
to calculate d take the second term
00:17:52
subtracted by the first term
00:17:54
7 minus 4 extreme or you could take the
00:17:57
third term subtracted by the second 10
00:17:59
minus 7 is 4.
00:18:02
in the case of f if you take 12 divided
00:18:04
by 3 you get 4.
00:18:05
48 divided by 12 you get 4. so that's
00:18:08
how you can calculate the common
00:18:10
difference or the common ratio
00:18:13
is by analyzing the second term with
00:18:15
respect to the first one
00:18:17
so since we have a common ratio
00:18:19
this is going to be geometric
00:18:24
for g if we subtract 18 by 12 we get a
00:18:28
common difference of positive six
00:18:30
24 minus 18
00:18:32
gives us the same common difference of
00:18:34
six
00:18:36
so this is going to be
00:18:38
arithmetic
00:18:40
so now let's put it all together let's
00:18:42
summarize the answers
00:18:45
so for part a what we have is a finite
00:18:47
arithmetic sequence
00:18:49
part b
00:18:50
this is an infinite geometric sequence
00:18:53
c
00:18:54
we have an infinite arithmetic series
00:18:58
d
00:18:59
is a finite geometric series
00:19:02
e is an infinite arithmetic sequence
00:19:06
f
00:19:06
is an infinite geometric series
00:19:09
g is a finite arithmetic series
00:19:13
so we have three columns of information
00:19:16
with two different possible choices
00:19:19
thus two to the third is eight which
00:19:21
means that we have eight different
00:19:22
possible combinations
00:19:25
right now i have seven out of the eight
00:19:27
different combinations the last one is a
00:19:30
finite
00:19:31
geometric sequence which i don't have
00:19:33
listed here
00:19:35
so now you know how to
00:19:36
identify whether you have a sequence or
00:19:39
series if it's a rhythmical geometric
00:19:41
and if it's finite or infinite
00:19:44
number two
00:19:45
write the first four terms of the
00:19:47
sequence defined by the formula a sub n
00:19:50
is equal to three n minus seven
00:19:55
so the first thing we're going to do is
00:19:56
find the first term
00:19:58
so we're going to replace n with one
00:20:01
so it's going to be three minus seven
00:20:04
which is negative 4.
00:20:06
and then we're going to repeat the
00:20:07
process we're going to find the second
00:20:08
term a sub 2.
00:20:11
so it's 3 times 2 minus 7
00:20:14
which is negative 1.
00:20:16
next we'll find a sub 3.
00:20:20
three times three is nine minus seven
00:20:22
that's two
00:20:23
and then the fourth term a sub four
00:20:26
that's going to be twelve minus seven
00:20:28
which is five
00:20:31
so we have a first term of negative four
00:20:33
then it's negative one
00:20:34
two
00:20:35
five
00:20:37
and then the sequence can continue
00:20:40
so the common difference in this problem
00:20:43
is positive three
00:20:45
going from negative one to two if you
00:20:47
add three you'll get two
00:20:48
and then two plus three is five
00:20:53
but this is the answer for the problem
00:20:55
so
00:20:55
this is
00:20:57
those are the first four terms of the
00:20:58
sequence
00:21:00
number three
00:21:01
write the next three terms of the
00:21:03
following arithmetic sequence
00:21:08
in order to find the next three terms we
00:21:10
need to determine the common difference
00:21:15
a simple way to find the common
00:21:16
difference is to subtract the second
00:21:18
term by the first term
00:21:21
22 minus 15 is 7.
00:21:26
now just to confirm we need to make sure
00:21:27
that
00:21:28
the difference between the third and the
00:21:30
second term is the same
00:21:32
29 minus 22 is also
00:21:35
seven
00:21:37
so we have a common difference of seven
00:21:40
so we could use that to find in the next
00:21:42
three terms
00:21:43
so 36 plus 7 is 43
00:21:46
43 plus 7 is 50
00:21:48
50 plus 7 is 57
00:21:51
so these are the next three terms of the
00:21:53
arithmetic sequence
00:21:56
here's a similar problem but presented
00:21:58
differently
00:21:59
write the first five terms of an
00:22:00
arithmetic sequence
00:22:03
given a one and d
00:22:06
so we know the first term is 29 and the
00:22:09
common difference is negative four
00:22:12
so this is all we need to write the
00:22:14
first five terms if the common
00:22:16
difference is negative four
00:22:17
then the next term is going to be 29
00:22:19
plus negative four which is 25
00:22:23
25 plus negative four or 25 minus 4 is
00:22:26
21
00:22:27
21 minus 4 is 17
00:22:29
17 minus 4 is 13.
00:22:32
so that's all we need to do in order to
00:22:34
write the first five terms
00:22:36
of the arithmetic sequence given this
00:22:38
information
00:22:40
number five
00:22:41
write the first five terms of the
00:22:43
sequence
00:22:44
defined by the following recursive
00:22:46
formulas
00:22:49
so let's start with the first one part a
00:22:52
so we're given the first term what are
00:22:54
the other terms
00:22:57
when dealing with recursive formulas
00:23:00
we need to realize is that you get the
00:23:02
next term by plugging in the previous
00:23:03
term
00:23:05
so let's say n is 2.
00:23:08
when n is two this is a sub two
00:23:12
and that's going to equal a sub n minus
00:23:15
one two minus one is one so this becomes
00:23:17
a sub one plus four
00:23:21
so the second term is going to be the
00:23:22
first term 3
00:23:24
plus 4 which is 7.
00:23:30
so we have 3 as the first term 7 as a
00:23:33
second term so now let's find the next
00:23:34
one
00:23:36
so let's plug in 3 for n so this becomes
00:23:39
a sub 3
00:23:42
the next one this becomes a sub 3 minus
00:23:46
1 or a sub 2
00:23:48
plus 4.
00:23:51
so this is seven
00:23:53
plus four which is eleven
00:23:55
at this point we can see that we have an
00:23:57
arithmetic sequence with a common
00:23:59
difference of four so to get the next
00:24:01
two terms we could just add four it's
00:24:03
going to be 15 and 19.
00:24:10
so that's it for part a
00:24:12
so when dealing with recursive formulas
00:24:13
just remember you get your next term by
00:24:15
using the previous term
00:24:18
now for part b it there's going to be a
00:24:20
little bit more work
00:24:22
so plugging in n equals 2
00:24:25
we have the second term
00:24:27
it's going to be 3 times the first term
00:24:30
plus 2.
00:24:32
the first term is two
00:24:34
so three times two is six plus two that
00:24:36
gives us eight
00:24:39
so now let's plug in n equals three
00:24:45
when n is 3 we have this equation a sub
00:24:48
3 is equal to 3 times a sub 2 plus 2.
00:24:53
so we're going to take 8 and plug it in
00:24:55
here to get the third term
00:24:58
so it's 3 times 8 plus 2
00:25:01
3 times 8 is 24 plus 2 that's
00:25:04
26.
00:25:08
now let's focus on the fourth term when
00:25:10
n is 4. so this is going to be a sub 4
00:25:12
is equal to 3 times
00:25:14
a sub 3 plus 2.
00:25:17
so now we're going to plug in 26 for a
00:25:19
sub 3.
00:25:21
so it's 3 times 26 plus 2.
00:25:28
3 times 26 is 78 plus 2 that's going to
00:25:31
be 80.
00:25:36
now let's focus on the fifth term
00:25:39
so a sub 5 is going to be 3 times a sub
00:25:43
4 plus 2
00:25:45
so that's 3 times 80 plus 2
00:25:49
3 times 8 is 24 so 3 times 80 is 240
00:25:53
plus 2 that's going to be 200
00:26:00
so the first five terms are 2
00:26:03
8
00:26:04
26
00:26:05
80
00:26:07
and
00:26:08
242 so this is neither
00:26:12
an arithmetic sequence nor is it a
00:26:14
geometric sequence
00:26:16
number six
00:26:18
write a general formula
00:26:20
or explicit formula which is the same
00:26:22
for the sequences shown below
00:26:25
in order to write a general formula or
00:26:27
an explicit formula
00:26:29
all we need is the first term and the
00:26:31
common difference
00:26:32
if it's an arithmetic sequence which for
00:26:34
part a
00:26:35
it definitely is
00:26:38
so if we subtract 14 by 8 we get 6 and
00:26:42
if we subtract 20 by 14 we get 6.
00:26:44
so we can see that the common difference
00:26:48
is positive 6
00:26:49
and the first term
00:26:51
is 8.
00:26:53
so the general formula is a sub n is
00:26:55
equal to a sub 1
00:26:57
plus n minus 1 times d
00:27:00
so all we need is the first term and the
00:27:02
common difference
00:27:04
and we can write a general formula or an
00:27:07
explicit formula
00:27:10
the first term is eight
00:27:13
d is six
00:27:14
now what we're going to do is we're
00:27:15
going to distribute six to n minus one
00:27:20
so we have six times n which is six n
00:27:23
and then this will be negative six
00:27:26
next we need to combine like terms
00:27:28
so eight plus negative six or eight
00:27:31
minus six that's going to be positive
00:27:32
two
00:27:34
so the general formula is six n
00:27:37
plus two
00:27:41
so if we were to plug in one
00:27:44
this will give us the first term eight
00:27:46
six times one plus two is eight
00:27:49
if we were to plug in four
00:27:50
it should give us the fourth term twenty
00:27:52
six
00:27:53
six times four is twenty four
00:27:55
plus two that's twenty six
00:27:59
so now that we have the explicit formula
00:28:01
for part a what about the sequence in
00:28:04
part b
00:28:05
what should we do if we have fractions
00:28:12
if you have a fraction like this or a
00:28:14
sequence of fractions and you need to
00:28:16
write an explicit formula
00:28:19
try to separate it into two different
00:28:20
sequences
00:28:22
notice that we have an arithmetic
00:28:24
sequence if we focus in the numerator
00:28:26
that sequence is
00:28:29
two
00:28:30
three
00:28:31
four
00:28:32
five and six
00:28:34
for the denominator we have the sequence
00:28:36
three five seven
00:28:37
nine eleven
00:28:40
so for the sequence on top the first
00:28:42
term is two and we can see that the
00:28:44
common difference is one
00:28:46
the numbers are increasing by one
00:28:49
so using the formula a sub n is equal to
00:28:51
a sub one plus n minus one times d
00:28:55
we have that a sub one is 2
00:28:58
and d is 1.
00:29:00
if you distribute 1 to n minus 1 you're
00:29:02
just going to get n minus 1.
00:29:05
so we can combine 2 and negative 1
00:29:08
which is positive 1.
00:29:10
so we get the formula n plus one
00:29:15
and you could check it when you plug in
00:29:17
one one plus one is two so the first
00:29:19
term is two
00:29:21
if you were to plug in
00:29:23
five
00:29:24
five plus one is six that will give you
00:29:26
the fifth term which is six
00:29:31
now let's focus on the sequence of the
00:29:33
denominators
00:29:35
the first term is three the common
00:29:37
difference we could see is two
00:29:39
five minus three is two
00:29:41
seven minus five is two
00:29:44
so using this formula again
00:29:46
we have a sub n is equal to a sub one a
00:29:49
sub one is three plus
00:29:51
n minus one times d d is two
00:29:56
so let's distribute two to n minus one
00:30:00
so that's gonna be two n minus two
00:30:03
and then let's combine like terms three
00:30:05
minus two is positive one
00:30:09
so a sub n is going to be two n
00:30:11
plus one
00:30:18
so if we want to calculate the first
00:30:20
term
00:30:21
we plug in one for n two times one is
00:30:23
two plus one it gives us
00:30:26
three
00:30:26
if we wanna calculate the fourth term
00:30:29
and it's four
00:30:30
two times four is eight plus one it
00:30:33
gives us nine
00:30:35
so you always want to double check your
00:30:36
work to make sure that you have the
00:30:38
right formula
00:30:40
so now let's put it all together
00:30:48
so we're going to write a sub n
00:30:51
and we're going to write it as a
00:30:52
fraction the sequence for the numerator
00:30:54
is n plus one
00:30:56
the sequence for the denominator is two
00:30:58
n plus one
00:31:02
so this right here
00:31:06
represents
00:31:08
the sequence
00:31:10
that corresponds to what we see in part
00:31:12
b
00:31:15
and we can test it out let's calculate
00:31:17
the value of the third term
00:31:20
so let's replace n with three it's going
00:31:22
to be three plus one
00:31:24
over two
00:31:25
times three plus one
00:31:27
three plus one is four
00:31:28
two times three is six plus one that's
00:31:30
seven so we get four over seven
00:31:33
if we wish to calculate the fifth term
00:31:35
it's going to be five plus one
00:31:38
over two
00:31:39
times five plus one
00:31:41
five plus one is six
00:31:42
two times five is ten plus one
00:31:45
that's 11.
00:31:49
and so anytime you have to write an
00:31:50
explicit formula given a sequence of
00:31:53
fractions
00:31:55
separate the numerator and the
00:31:56
denominator into two different sequences
00:31:58
hopefully they're both arithmetic
00:32:01
if it's geometric you may have to look
00:32:02
at another video that i'm going to make
00:32:03
soon on geometric sequences
00:32:06
but
00:32:07
break it up into two separate sequences
00:32:09
and then write the formulas that way and
00:32:11
then put the two formulas in a fraction
00:32:13
and that's how you can get the answer
00:32:15
number seven
00:32:16
write a formula for the nth term of the
00:32:19
arithmetic sequences shown below
00:32:24
so writing a formula for the f term is
00:32:26
basically the same as writing a general
00:32:27
formula for the sequence or an explicit
00:32:29
formula
00:32:31
so we need to identify the first term
00:32:33
which we could see as 5
00:32:35
and the common difference
00:32:38
14 minus 5 is 9
00:32:40
23 minus 14 is 9 as well
00:32:44
so once we have these two we can write
00:32:47
the general formula
00:32:52
so let's replace the first term a sub 1
00:32:54
with 5.
00:32:56
and let's replace d with nine
00:32:59
now let's distribute nine to n minus one
00:33:05
so we're gonna have nine n minus nine
00:33:08
next let's combine like terms
00:33:13
so it's going to be 9n and then 5 minus
00:33:16
9 is negative 4.
00:33:19
so this
00:33:20
is the formula for the nth term of the
00:33:23
sequence
00:33:32
now let's do the same for part b
00:33:35
so the first term is 150
00:33:38
the common difference is going to be 143
00:33:41
minus 150
00:33:42
which is negative seven to confirm that
00:33:45
if you subtract 136 by 143 you also get
00:33:49
negative seven
00:33:53
now let's plug it into this formula to
00:33:55
write the general equation
00:33:59
so a sub n is going to be 150
00:34:01
plus
00:34:02
n minus 1 times d which is negative
00:34:05
seven
00:34:06
so let's distribute negative seven to n
00:34:08
minus one
00:34:12
so it's going to be 150 minus seven n
00:34:15
and then negative seven times negative
00:34:17
one that's going to be positive 7.
00:34:20
so a sub n is going to be negative 7n
00:34:24
plus 157
00:34:27
or you could just write it as
00:34:30
157
00:34:32
minus 7n
00:34:36
so that is the formula for the nth term
00:34:39
of the arithmetic sequence
00:34:44
now let's move on to part b
00:34:46
calculate the value of the tenth term
00:34:49
of the sequence
00:34:52
so we're looking for a sub 10. so let's
00:34:54
plug in 10 into this equation
00:34:56
so it's gonna be nine
00:34:57
times ten
00:34:59
minus four
00:35:00
nine times ten is ninety
00:35:02
ninety minus four is eighty-six
00:35:05
so that is the tenth term
00:35:08
of the sequence in part a
00:35:10
for part b
00:35:11
the tenth term is going to be 157
00:35:15
minus seven times ten
00:35:18
seven times ten is seventy one fifty
00:35:20
seven
00:35:22
minus seventy
00:35:23
is going to be eighty seven
00:35:30
now let's move on to part c
00:35:31
find the sum of the first 10 terms
00:35:37
so in order to find the sum we need to
00:35:39
use this formula
00:35:43
s sub n is equal to the first term plus
00:35:45
the last term divided by 2
00:35:48
times the number of terms
00:35:51
so if we want to find the sum of the
00:35:52
first 10 terms we need a sub 1 which we
00:35:55
know it's 5.
00:35:58
a sub n n is 10 so that's a sub 10
00:36:02
the tenth term is 86
00:36:04
divided by 2
00:36:06
times the number of terms which is 10.
00:36:10
5 plus 86 is 91. 91 divided by 2
00:36:14
gives us an average of 45.5 of the first
00:36:17
and last number
00:36:19
and then times 10
00:36:21
we get a total sum of 455.
00:36:25
so that is the sum of the first 10 terms
00:36:28
of this sequence
00:36:31
now for part b
00:36:33
we're going to do the same thing
00:36:34
calculate s sub 10
00:36:36
the first term a sub 1 is 150
00:36:40
the tenth term is eighty seven
00:36:44
divided by two times the number of terms
00:36:47
which is ten
00:36:50
one fifty plus eighty seven that's two
00:36:52
thirty seven
00:36:53
divided by 2 that's
00:36:55
118.5
00:36:57
times 10
00:36:58
we get a sum of 11.85
00:37:05
so now you know how to calculate the
00:37:06
value of the m term and you also know
00:37:09
how to find the sum
00:37:10
of
00:37:11
a series
00:37:14
number eight
00:37:15
find the sum of the first 300 natural
00:37:17
numbers
00:37:20
so how can we do this
00:37:22
the best thing we can do right now is
00:37:23
write a series
00:37:25
zero is not a natural number but one is
00:37:28
so if we write a list one plus two plus
00:37:31
three
00:37:32
and this is going to keep on going
00:37:35
all the way to 300
00:37:39
so to find the sum of a partial series
00:37:41
we need to use this equation s sub n is
00:37:43
equal to a sub 1
00:37:45
plus a sub n over 2 times n
00:37:50
now let's write down what we know
00:37:52
we know that a sub 1 the first term is
00:37:55
one
00:37:56
we know n
00:37:57
is 300
00:37:59
if this is the first term this is the
00:38:00
second term this is the third term this
00:38:03
must be the 300th term
00:38:06
so we know n is 300 and
00:38:08
a sub n or a sub 300 is 300
00:38:12
so we have everything that we need to
00:38:13
calculate the sum of the first 300 terms
00:38:17
so it's a sub 1 which is 1 plus a sub n
00:38:20
which is 300
00:38:22
over 2
00:38:23
times the number of terms which is 300
00:38:27
so it's going to be 301 divided by 2
00:38:30
times 300
00:38:32
and that's
00:38:33
45
00:38:35
150.
00:38:37
so that's how we can calculate the sum
00:38:39
of the first 300
00:38:41
natural numbers
00:38:43
in this series
00:38:46
number nine
00:38:47
calculate the sum of all even numbers
00:38:49
from two to one hundred inclusive
00:38:52
so let's write a series
00:38:55
two is even three is odd so the next
00:38:57
even number is four
00:38:59
and then six and then eight
00:39:02
all the way to one hundred
00:39:05
so we have the first term
00:39:07
the second term is four
00:39:09
the third term is sixty
00:39:12
one hundred is likely to be the 50th
00:39:14
term but let's confirm it
00:39:16
so what we need to do is calculate n and
00:39:18
make sure it's 50 and not 49 or 51.
00:39:23
so we're going to use this equation to
00:39:24
calculate the value of n
00:39:30
so a sub n is a hundred
00:39:34
let's replace that with a hundred
00:39:37
a sub one
00:39:38
is two
00:39:43
the common difference
00:39:46
we see four minus two is two six minus
00:39:49
four is two so the common difference
00:39:52
is two in this example
00:39:54
and our goal is to solve for n
00:39:57
so let's begin by subtracting both sides
00:39:59
by two
00:40:01
a hundred minus two is ninety-eight
00:40:06
and this is going to equal two times n
00:40:07
minus one
00:40:09
next we're going to divide both sides by
00:40:11
two
00:40:14
98 divided by two is 49
00:40:17
so we have 49 is equal to n minus one
00:40:20
and then we're going to add one to both
00:40:22
sides so n is 49 plus 1 which is 50.
00:40:28
so that means that
00:40:30
100 is indeed the 50th term so we know
00:40:33
that n
00:40:34
is 50.
00:40:36
so now we have everything that we need
00:40:37
in order to calculate the sum of the
00:40:39
first 50 terms
00:40:41
so let's begin by writing out the
00:40:43
formula first
00:40:48
so the sum of the first 50 of terms is
00:40:50
going to be the first term which is 2
00:40:53
plus a sub 50 the last term which is 100
00:40:56
divided by 2
00:40:58
times n
00:40:59
which is 50.
00:41:02
so 2 plus 100 that's 102 divided by 2
00:41:05
that's 51.
00:41:06
51 times 50
00:41:08
is 2550.
00:41:12
so that is the sum
00:41:14
of all of the even numbers from 2 to 100
00:41:16
inclusive
00:41:17
try this one determine the sum of all
00:41:20
odd integers from 20 to 76
00:41:24
20 is even but the next number 21 is odd
00:41:29
and then 23 25 27
00:41:32
all of that are odd numbers up until 75
00:41:37
so a sub 1
00:41:39
is 21 in this problem
00:41:44
the last number a sub n
00:41:47
is 75
00:41:50
and we know the common difference is two
00:41:52
because the numbers are increased by two
00:41:56
what we need to calculate is the value
00:41:58
of n
00:41:59
once we could find n then we could find
00:42:01
the sum from
00:42:02
21 to 75.
00:42:06
so what is the value of n
00:42:09
so we need to use
00:42:11
the general formula for
00:42:13
an arithmetic sequence
00:42:15
so a sub n is 75 a sub 1 is 21
00:42:20
and the common difference is 2.
00:42:23
so let's subtract both sides by 21
00:42:27
75 minus 21 this is going to be 54.
00:42:35
dividing both sides by 2.
00:42:40
54 divided by 2 is 27. so we get 27 is n
00:42:44
minus 1
00:42:45
and then we're going to add 1 to both
00:42:47
sides
00:42:49
so n is 28
00:42:52
so a sub 28 is 75
00:42:56
75 is the 28th term in the sequence
00:43:00
so now we need to find the sum of the
00:43:03
first 28 terms
00:43:04
it's going to be a sub 1 the first term
00:43:07
plus the last term or the 28th term
00:43:09
which is 75
00:43:11
divided by 2
00:43:13
times the number of terms which is
00:43:15
28
00:43:21
21 plus 75 that's 96
00:43:25
divided by 2 that's 48 so 48 is the
00:43:28
average of the first and the last term
00:43:31
so 48 times 28
00:43:34
that's 1
00:43:35
344.
00:43:37
so that is the sum of the first 28 terms
00:44:03
you
