Question

Find the 14th term of the arthmetic sequence x-1, 7x -7 , 13x-13,....

Answer 1

Since they're in arithmetic progression, there is a constant difference between consecutive terms of

`(7x - 7) - (x - 1) = 6x - 6`

Recursively, the sequence is given by

`\begin{cases} a_1 = x-1 \\ a_n = a_(n-1) + 6x - 6 & \text{for }n\ge2\end{cases}`

By substitution, we have

`a_2 = a_1 + 6x-6`

`a_3 = a_2 + 6x-6 = a_1 + 2(6x-6)`

`a_4 = a_3 + 6x-6 = a_1 + 3(6x-6)`

and so on, up to the 14th term

`a_(14) = a_1 + 13(6x-6)`

`a_(14) = (x-1) + 78 (x - 1)`

`a_(17) = 79 (x - 1) = \boxed{79x-79}`