Find the geometric sequence where 𝑇 one, the first term, equals 500 and 𝑇 three, the third term, equals 125.
In any geometric sequence, we let 𝑎 be the first term and 𝑟 be the common ratio. The common ratio is the number that we multiply to get from the first term to the second term, also from the second term to the third term, and so on. This means that when the first term is 𝑎, the second term will be 𝑎 multiplied by 𝑟. The third term will be 𝑎 multiplied by 𝑟 squared. This pattern leads us to the general formula for the 𝑛th term, or 𝑇𝑛, of 𝑎 multiplied by 𝑟 to the power of 𝑛 minus one.
In this specific question, we were told that the first term, 𝑇 one, was equal to 500. This means that 𝑎 is equal to 500. We were also told that the third term, 𝑇 three, was equal to 125. Therefore, 𝑎𝑟 squared is equal to 125. As we already know what the first term is, our next step is to work out the common ratio. If we substitute 𝑎 equals 500 into equation two, we end up with the equation 500 𝑟 squared equals 125. Dividing both sides of the equation by 500 gives us 𝑟 squared equals 0.25, as 125 divided by 500 is a quarter or 0.25. Square rooting both sides of this equation gives us two possible answers for 𝑟. 𝑟 could be positive 0.5 or negative 0.5. This is because positive 0.5 multiplied by positive 0.5 is 0.25. Similarly, negative 0.5 multiplied by negative 0.5 is also equal to 0.25.
As we found two possible values of 𝑟, there are two possible sequences, with a first term of 500 and a third term of 125. Firstly, when 𝑟 equals positive 0.5, the first three terms of our geometric sequence are 500, 250, and 125. This is because 500 multiplied by 0.5 is 250. 250 multiplied by 0.5 is 125. The second possible sequence, when 𝑟 is equal to negative 0.5, has first three terms: 500, negative 250, 125. This is because 500 multiplied by negative 0.5 is negative 250. Negative 250 multiplied by negative 0.5 is positive 125. This confirms that there are two possible geometric sequences with first term 500 and third term 125.