Verify that each of the following is an $AP$,and then write its next three terms.
$0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots$
$0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots$
(N/A) Here,$a_{1}=0, a_{2}=\frac{1}{4}, a_{3}=\frac{1}{2}$ and $a_{4}=\frac{3}{4}$.
Calculate the common difference $d$:
$a_{2}-a_{1} = \frac{1}{4} - 0 = \frac{1}{4}$
$a_{3}-a_{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$
$a_{4}-a_{3} = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}$
Since the difference between consecutive terms is constant $(d = \frac{1}{4})$,the given sequence forms an $AP$.
The next three terms are:
$a_{5} = a_{4} + d = \frac{3}{4} + \frac{1}{4} = 1$
$a_{6} = a_{5} + d = 1 + \frac{1}{4} = \frac{5}{4}$
$a_{7} = a_{6} + d = \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2}$
Calculate the common difference $d$:
$a_{2}-a_{1} = \frac{1}{4} - 0 = \frac{1}{4}$
$a_{3}-a_{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$
$a_{4}-a_{3} = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}$
Since the difference between consecutive terms is constant $(d = \frac{1}{4})$,the given sequence forms an $AP$.
The next three terms are:
$a_{5} = a_{4} + d = \frac{3}{4} + \frac{1}{4} = 1$
$a_{6} = a_{5} + d = 1 + \frac{1}{4} = \frac{5}{4}$
$a_{7} = a_{6} + d = \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2}$
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