Show that $a_{1}, a_{2}, a_{3}, . . . . . . . a_{n}$ form an AP where $a_{n} = 3 + 4 n$

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Solution

Given:-
$a_{n} = 3 + 4 n$
Taking $n = 1$ ,
$a_{1} = 3 + 4 \times 1 = 7$
$a_{2} = 3 + 4 \times 2 = 11$
$a_{3} = 3 + 4 \times 3 = 15$
Therefore, the series is-
$7, 11, 15, . . . . . . .$
Now,
$a_{2} - a_{1} = 11 - 7 = 4$
$a_{3} - a_{2} = 15 - 11 = 4$
Since the difference is same, i.e., $4$ , thus the above series form an A.P.

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