Verify that each of the following lists of numbers is an A.P., and then write its next three terms:

(i) 0, $\dfrac{1}{4}, \dfrac{1}{2}, \dfrac{3}{4}, …$

(ii) 5, $\dfrac{14}{3}, \dfrac{13}{3}, 4, …$

(i) Here,

$a$2 - a1 = \dfrac{1}{4} - 0 = \dfrac{1}{4},\\[1em] a3 - a2 = \dfrac{1}{2} - \dfrac{1}{4} = \dfrac{2 - 1}{4} = \dfrac{1}{4}. \\[1em]

i.e. any term - preceding term = $\dfrac{1}{4}$ = fixed number.

Hence, the given list of numbers forms an A.P.

For the next terms, we have,

$a$5 = a4 + d = \dfrac{3}{4} + \dfrac{1}{4} = \dfrac{4}{4} = 1, \\[1em] a6 = a5 + d = 1 + \dfrac{1}{4} = \dfrac{4 + 1}{4} = \dfrac{5}{4}, \\[1em] a7 = a6 + d = \dfrac{5}{4} + \dfrac{1}{4} = \dfrac{6}{4} = \dfrac{3}{2}.

So, the next three terms are $1, \dfrac{5}{4}, \dfrac{3}{2}.$

(ii) Here,

$a$2 - a1 = \dfrac{14}{3} - 5 = \dfrac{14 - 15}{3} = -\dfrac{1}{3},\\[1em] a3 - a2 = \dfrac{13}{3} - \dfrac{14}{3} = \dfrac{13 - 14}{3} = -\dfrac{1}{3}. \\[1em]

i.e. any term - preceding term = $-\dfrac{1}{3}$ = fixed number.
Hence, the given list of numbers forms an A.P.

For the next terms, we have,

$a$5 = a4 + d = 4 + \Big(-\dfrac{1}{3}\Big) = 4 - \dfrac{1}{3} = \dfrac{12 - 1}{3} = \dfrac{11}{3}, \\[1em] a6 = a5 + d = \dfrac{11}{3} + \Big(-\dfrac{1}{3}\Big) = \dfrac{11}{3} - \dfrac{1}{3} = \dfrac{10}{3}, \\[1em] a7 = a6 + d = \dfrac{10}{3} + \Big(-\dfrac{1}{3}\Big) = \dfrac{10}{3} - \dfrac{1}{3} = \dfrac{9}{3} = 3.

So, the next three terms are $\dfrac{11}{3}, \dfrac{10}{3}, 3.$