Math, asked by SidMhatre, 1 year ago

exercise 5.1 1st 2nd

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Answers

Answered by fganesh2231
0
Let the initial volume of air in a cylinder be V lit. In each stroke,
the vacuum pump removes ¼ of air remaining in the cylinder at a time.
In other words, after every stroke, only 1 −
1
4
=
3
4
푡ℎ part of air will remain.
Therefore, volumes will be 푉,(3푉
4
),(3푉
4
)
2
, (3푉
4
)
3

Clearly, it can be observed that the adjacent terms of this series do not
have the same difference between them. Therefore, this is not an A.P.
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Answered by sowmya36
0
Amount of air present in the cylinder ='x'=x=a1. Amount of air present in the cylinder after 1 time removal of air by the vacccum pump=x-1/4(x) = x-x/4 = 4x-x/4 = 3x/4 = a2. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Amount of air present in the cylinder after 2 times removal of the air by the vacccum pump. 3x/4-1/4(3x/4) = 3x/4-3x/16 = 12x-3x/16 = 9/16=(3/4)square x=a3. •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• x , 3x/4 , 9x/16. a2-a1 = 3x/4-x = 3x-4x/4 = -x/4. a3-a2 = 9x/16-3x/4 = 9x/16-12x/16 = 9x-12x/16=-3x/16. a2-a1 not equal to a3-a2. It not form an AP

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