Video Transcript
In a geometric sequence, the first
term is 𝑎 and the common ratio is 𝑟. Find the sum of the first three
terms of a geometric sequence with 𝑎 equal to 328 and 𝑟 equal to one-quarter.
There are a couple of ways we could
approach this problem. We are told that the first term 𝑎
is equal to 328. The second term of a geometric
sequence is equal to 𝑎 multiplied by 𝑟. As 𝑟 is equal to one-quarter, we
need to multiply 328 by one-quarter. This is the same as dividing 328 by
four, which gives us 82. The third term is equal to 𝑎𝑟
squared. This means that we need to multiply
328 by one-quarter squared or multiply 82 by one-quarter. This is equal to 41 over two or
20.5.
We need to calculate the sum of
these first three terms. We need to add 328, 82, and 41 over
two. This is equal to 861 over two. The sum of the first three terms of
the geometric sequence is 861 over two or 430.5.
An alternative method here would be
to use the formula for the sum of the first 𝑛 terms of a geometric sequence. This states that 𝑆 of 𝑛 is equal
to 𝑎 multiplied but one minus 𝑟 to the 𝑛th power divided by one minus 𝑟. Substituting in our values of 𝑎,
𝑟, and 𝑛 equals three gives us 328 multiplied by one minus one-quarter cubed
divided by one minus a quarter.
Typing this into the calculator, we
get the same answer of 861 over two. This is the sum of the first three
terms, where 𝑎 is equal to 328 and 𝑟 is equal to one-quarter.
