The 𝑛th term in a sequence is given by 𝑎 sub 𝑛 is equal to 𝑎 plus 𝑛𝑏. Given that 𝑎 sub one is equal to 65 and 𝑎 sub three is equal to three, find the values of 𝑎 and 𝑏.
In this question, we are given the general rule 𝑎 sub 𝑛 is equal to 𝑎 plus 𝑛𝑏. As 𝑎 sub one is equal to 65, we can substitute 𝑛 equals one into the general formula. This gives us 𝑎 plus one 𝑏 is equal to 65. As one 𝑏 is just 𝑏, this can be rewritten as 𝑎 plus 𝑏 is equal to 65. We will call this equation one.
We are also told that 𝑎 sub three, or the third term, is equal to three. In this case, 𝑛 is equal to three. Therefore, 𝑎 plus three 𝑏 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 𝑎’s from the equations by subtracting equation one from equation two. 𝑎 minus 𝑎 is equal to zero. Three 𝑏 minus 𝑏 is equal to two 𝑏. Three minus 65 is equal to negative 62. Dividing both sides of this equation by two gives us 𝑏 is equal to negative 31. We can now substitute this value into equation one or equation two to calculate the value of 𝑎.
Replacing 𝑏 with negative 31 in equation one gives us 𝑎 plus negative 31 equals 65. Adding 31 to both sides of this equation gives us 𝑎 is 96. The values of 𝑎 and 𝑏 are 96 and negative 31, respectively. This means that the 𝑛th term in the sequence is given by 𝑎 sub 𝑛 is equal to 96 minus 31𝑛.