Question

Question #067a8

Answer 1

`a_n = a_1+(n-1)d`

Explanation:

Given an arithmetic progression with initial term `a_1` and a difference `d` between successive terms, that is, `a_k = a_(k-1)+d`, then the general term for the `n^"th"` term is given by

`a_n = a_1+(n-1)d`

Proof: (By induction)

Base case: For `n=1`, we have `a_1 = a_1+0d = a_1(n-1)d`

Inductive hypothesis: Suppose that for some integer `k>0`, we have `a_k = a_1+(k-1)d`.

Induction step: We wish to know that `a_(k+1) = a_1+((k+1)-1)d`. Indeed,

`a_(k+1) = a_k + d`

`=a_1+(k-1)d + d` (by the inductive hypothesis)

`=a_1 + kd`

`=a_1+((k+1)-1)d`

Thus, by induction, `a_n = a_1+(n-1)d` for all integers `n>=1`.
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