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Questions
In the following situation, involved make an arithmetic progression? and why?
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.
In the following situation, the sequence of number formed will form an A.P.?
The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of their remaining in the cylinder.
Solution 1
Let the quantity of air in the cylinder be 1.
T1 = 1
Air removed = `1/4`
`"T"_2 = 1 - 1/4 = (4 - 1)/4 = 3/4`
Air removed = `3/4 xx 1/4 = 3/16`
`"T"_3 = 3/4 - 3/16 = (12 - 3)/16 = 9/16`
Air removed = `9/16 xx 1/4 = 9/64`
`"T"_4 = 9/16 - 9/64 = (36 - 9)/64 = 27/64`
Series: `1, 3/4, 9/16, 27/64`
`"d"_1 = 3/4 - 1 = (-1)/4`
`"d"_2 = 9/16 - 3/4 = (-3)/16`
Here the common difference is not the same hence it is A.P. Not there.
Solution 2
Here, let us take the initial amount of air present in the cylinder as 100 units.
So,
Amount left after vacuum pump removes air for 1st time = `100 - (1/4) 100`
= 100 - 25
= 75
Amount left after vacuum pump removes air for 2nd time = `75 - (1/4)75`
= 75 - 18.75
= 56.25
Amount left after vacuum pump removes air for 3rd time = `56.25 - (1/4) 56.25`
= 56.25 - 14.06
= 42.19
Thus, the amount left in the cylinder at various stages is 100, 75, 56, 25, 42, 19
Now, for a sequence to be an A.P., the difference between adjacent terms should be equal.
Here
`a_1 - a = 75 - 100`
= -25
Also
`a_2 - a_1 = 56.25 - 75`
= -18.75
Since `a_1- a != a_2 - a_1`
The sequence is not an A.P.
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