The common difference of the A.P. 13,1−3b3,1−6b3, ... is

Question:

The common difference of the A.P. $\frac{1}{3}, \frac{1-3 b}{3}, \frac{1-6 b}{3}, \ldots$ is

(a) $\frac{1}{3}$

(b) $-\frac{1}{3}$

(c) $-b$

(d) $b$

Solution:

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

The given A.P. is $\frac{1}{3}, \frac{1-3 b}{3}, \frac{1-6 b}{3}, \ldots$

Common difference = d = Second term − First term

$=\frac{1-3 b}{3}-\frac{1}{3}$

$=\frac{-3 b}{3}=-b$

​Hence, the correct option is (c).

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