Question
Justify whether it is true to say that the sequence, having following $n^{th}$​​​​​​​ term is an A.P.
$a_n = 2n - 1.$

Answer

Yes, here $a_n = 2n - 1$
Put $n = 1, a_1 = 2(1) - 1 = 1$
Put $n = 2, a_2 = 2(2) - 1 = 3$
Put $n = 3, a_3 = 2(3) - 1 = 5$
Put $n = 4, a_4 = 2(8) - 1 = 7$
List of numbers becomes $1, 3, 5, 7, .....$
Here,
$a_2 - a_1 = 3 - 1 = 2$
$a_3 - a_2 = 5 - 3 = 2$
$a_4 - a_3 = 7 - 5 = 2$
$a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = .....$
Hence, $2n - 1$ is the $n^{th}$  term of A.P.

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