Question
The common difference of the A.P. is $$\frac{1}{{2q}},$$ $$\frac{{1 - 2q}}{{2q}},$$ $$\frac{{1 - 4q}}{{2q}},$$ . . . . is
Answer: Option A
$$\eqalign{
& {\text{A}}{\text{.P}}{\text{.}}\,{\text{is}}\,\frac{1}{{2q}},\,\frac{{1 - 2q}}{{2q}},\,\frac{{1 - 4q}}{{2q}},.... \cr
& \Rightarrow \frac{1}{{2q}},\,\left( {\frac{1}{{2q}} - 1} \right),\,\left( {\frac{1}{{2q}} - 2} \right),\,.... \cr
& {\text{Clearly}}\,d = \left( {\frac{1}{{2q}} - 1} \right) - \frac{1}{{2q}} \cr
& = \frac{1}{{2q}} - 1 - \frac{1}{{2q}} \cr
& = - 1 \cr} $$
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$$\eqalign{
& {\text{A}}{\text{.P}}{\text{.}}\,{\text{is}}\,\frac{1}{{2q}},\,\frac{{1 - 2q}}{{2q}},\,\frac{{1 - 4q}}{{2q}},.... \cr
& \Rightarrow \frac{1}{{2q}},\,\left( {\frac{1}{{2q}} - 1} \right),\,\left( {\frac{1}{{2q}} - 2} \right),\,.... \cr
& {\text{Clearly}}\,d = \left( {\frac{1}{{2q}} - 1} \right) - \frac{1}{{2q}} \cr
& = \frac{1}{{2q}} - 1 - \frac{1}{{2q}} \cr
& = - 1 \cr} $$
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