Video Transcript
Find the possible infinite
geometric series where the sum is negative 21 and the sum of the first and second
terms is negative 35 over three.
Let’s begin by recalling what we
know about a geometric series or sequence. The first term of any geometric
series is denoted by the letter 𝑎. The common ratio is equal to
𝑟. This is the number that we multiply
by to get from the first to the second term, the second to the third, and so on. A geometric sequence, therefore,
has terms 𝑎, 𝑎𝑟, 𝑎𝑟 squared, and so on. And a geometric series can,
therefore, be written as 𝑎 plus 𝑎𝑟 plus 𝑎𝑟 squared, et cetera.
The sum of the first 𝑛 terms in a
geometric series is equal to 𝑎 multiplied by one minus 𝑟 to the power of 𝑛
divided by one minus 𝑟. The sum to ∞ of a geometric series
is equal to 𝑎 divided by one minus 𝑟. This is only valid when the
absolute value or modulus of 𝑟 is less than one.
We are told in this question that
the sum to ∞ is equal to negative 21. This means that 𝑎 divided by one
minus 𝑟 is equal to negative 21. Multiplying both sides of this
equation by one minus 𝑟 gives us 𝑎 is equal to negative 21 multiplied by one minus
𝑟. Distributing the parentheses gives
us 𝑎 is equal to negative 21 plus 21𝑟. This can be rewritten as 21𝑟 minus
21. We will call this equation one.
We’re also told that the sum of the
first and second terms is negative 35 over three. We could use the 𝑠 of 𝑛 formula
for the sum of the first two terms. However, it is easier to write the
first term as 𝑎 and the second term as 𝑎𝑟. 𝑎 plus 𝑎𝑟 is equal to negative
35 over three. Multiplying both sides of this
equation by three gives us three 𝑎 plus three 𝑎𝑟 is equal to negative 35. We could factor out terms on the
left-hand side. However, at this stage, we will
call this equation two.
Instead of factoring, we can now
substitute equation one into equation two. This gives us three multiplied by
21𝑟 minus 21 plus three 𝑟 multiplied by 21𝑟 minus 21 is equal to negative 35. We can then distribute both of the
parentheses. Our equation simplifies to 63𝑟
minus 63 plus 63𝑟 squared minus 63𝑟 is equal to negative 35. 63𝑟 minus 63𝑟 is equal to
zero. Our equation simplifies to 63𝑟
squared minus 63 is equal to negative 35.
We will now clear some room to
continue our working. Adding 63 to both sides of the
equation gives us 63𝑟 squared is equal to 28. Dividing both sides by 63 gives us
𝑟 squared is equal to 28 over 63. Both the numerator and denominator
are divisible by seven. This means that 𝑟 squared is equal
to four over nine or four-ninths.
Our next step is to take the square
root of both sides. So 𝑟 is equal to the square root
of four-ninths. The answer to this square root can
be positive or negative. As the square root of four is two
and the square root of nine is three, 𝑟 could be equal to two-thirds or negative
two-thirds.
We recall that equation one was 𝑎
equals 21𝑟 minus 21. We can now substitute each of our
values of 𝑟 into this equation. 𝑎 is equal to 21 multiplied by
two-thirds minus 21, or 𝑎 is equal to 21 multiplied by negative two-thirds minus
21. Two-thirds of 21 is equal to 14,
and 14 minus 21 is equal to negative seven. Multiplying 21 by negative
two-thirds gives us negative 14. Negative 14 minus 21 is equal to
negative 35. The two possible values of 𝑎 are
negative seven and negative 35.
We can now use these numbers to
write out the two possible infinite geometric series. The first option has a first term
of negative seven. Multiplying this by two-thirds
gives us a second term of negative 14 over three. Multiplying this again by
two-thirds gives us a third term of negative 28 over nine.
The first three terms in the
sequence would be negative seven, negative 14 over three, and negative 28 over
nine. As we’re looking for the series, we
need to write this as the sum of the three numbers, negative seven plus negative 14
over three plus negative 28 over nine. This can be rewritten as negative
seven minus 14 over three minus 28 over nine.
The second possible series had a
first term of negative 35. Multiplying this by negative
two-thirds gives us 70 over three. Multiplying this by negative
two-thirds again gives us negative 140 over nine.
The geometric series is equal to
negative 35 plus 70 over three plus negative 140 over nine. Once again, this can be rewritten,
giving us negative 35 plus 70 over three minus 140 over nine. These are the two possible infinite
geometric series where the sum is negative 21 and the sum of the first and second
terms is negative 35 over three.
