Question

the amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
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Answer 1

Answer:

Let

$\\ \qquad \sf \bullet \: \: a_1 = x \\ \\ \\ \qquad \sf \bullet \: \: a_2 = x - \dfrac{1}{4} \: x = \dfrac{3x - x}{4} = \dfrac{3}{4} \: x \\ \\ \\ \qquad \sf \bullet \: \: a_3 = \dfrac{3}{4} \: x - \dfrac{1}{4} \: x \: \bigg( \dfrac{3}{4} \: x \bigg) \\ \\ \\ \implies \sf \: \dfrac{3}{4} \: x - \dfrac{3}{16} \: x \\ \\ \\ \implies \sf \: \dfrac{12x - 3x}{16} \\ \\ \\ \implies \sf \: \frac{9}{16x}$

Now,

$\\ \\ \sf \: a_2 - a_1 = \dfrac{3}{4} \: x - x = \dfrac{ - 1}{4} \: x \\ \\ \\ \sf a_3 - a_2 = \dfrac{9}{16} \: x - \dfrac{3}{4} \: x \\ \\ \\ \implies \sf \: \dfrac{9x - 12x}{16} \\ \\ \\ \implies \sf \: \dfrac{ - 3}{16} \\ \\ \\ \large{ \underline{ \boxed{ \sf{ \green{ \: \because \: a_2 - a_1 \: \neq \: a_3 - a_2}}}}}$

i.e, common difference is not same. Therefore, it is not in AP.