### Video Transcript

The πth term in a sequence is given by π sub π is equal to π plus ππ. Given that π sub one is equal to 65 and π sub three is equal to three, find the values of π and π.

In this question, we are given the general rule π sub π is equal to π plus ππ. As π sub one is equal to 65, we can substitute π equals one into the general formula. This gives us π plus one π is equal to 65. As one π is just π, this can be rewritten as π plus π is equal to 65. We will call this equation one.

We are also told that π sub three, or the third term, is equal to three. In this case, π is equal to three. Therefore, π plus three π equals three. We will call this equation two. And we now have a pair of simultaneous equations that we can solve by elimination or substitution.

We can eliminate the πβs from the equations by subtracting equation one from equation two. π minus π is equal to zero. Three π minus π is equal to two π. Three minus 65 is equal to negative 62. Dividing both sides of this equation by two gives us π is equal to negative 31. We can now substitute this value into equation one or equation two to calculate the value of π.

Replacing π with negative 31 in equation one gives us π plus negative 31 equals 65. Adding 31 to both sides of this equation gives us π is 96. The values of π and π are 96 and negative 31, respectively. This means that the πth term in the sequence is given by π sub π is equal to 96 minus 31π.