Let the first term be ( a ) and common difference ( d ).

[ a_k = a + (k-1)d, \quad a_m = a + (m-1)d ]

Then [ a_k - a_m = (k - m)d \implies d = \dfrac{a_k - a_m}{k - m} ]

Now [ a_n = a + (n-1)d = \bigl[ a + (k-1)d \bigr] + (n - k)d = a_k + (n - k)d ]

[ a_n = a_k + (n - k)\left( \dfrac{a_k - a_m}{k - m} \right) ]

Proved.