Find the geometric means of the
sequence whose first term is two and whose last term is 4802.
Remember, 𝑛 geometric means
between two numbers are the 𝑛 terms of a geometric sequence between the two given
numbers. So, we’re going to need to identify
the sequence whose first term is two and whose last term is 4802 and where there are
exactly three terms between them. So, we use the formula for the 𝑛th
term of a geometric sequence with first term 𝑎 sub one and common ratio 𝑟. It’s 𝑎 sub 𝑛 equals 𝑎 sub one
times 𝑟 to the power of 𝑛 minus one.
The first term in our sequence is
two; then the fifth term is 4802. But then using our formula, we can
express that in terms of 𝑟; it’s two times 𝑟 to the power of five minus one or two
times 𝑟 to the fourth power. So, we now have an equation which
we can solve to find the value of 𝑟.
We begin by dividing both sides of
this equation by two. So, 𝑟 to the fourth power is
2401. Then, we can find the positive and
negative fourth root of 2401. And remember, we do this because
four is an even exponent or power, and this gives us a value for 𝑟 as positive or
negative seven. And so, in fact, there are two
possible sequences we’re interested in. The first is when 𝑟 is equal to
seven. Beginning with our first term two,
we multiply by seven each time. And this gives us the sequence two,
14, 98, 686, and 4802. And if 𝑟 is negative seven, we
change the sign of the 14 and the 686. So, the geometric means of this
sequence are either 14, 98, 686 or negative 14, 98, and negative 686.