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Q1(ii):
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (ii) The amount of air present in a cylinder when a vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at a time.

Solution :

Given: A cylinder contains an initial amount of air. A vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at each stroke.

To Find: Determine whether the sequence of the amounts of air remaining in the cylinder after each stroke forms an Arithmetic Progression (AP).

Step 1: Defining the Variables
Let the initial amount of air present in the cylinder be $V$ units.
Let $a_1$ be the amount of air after the 0th stroke (initial state).
Let $a_2$ be the amount of air after the 1st stroke.
Let $a_3$ be the amount of air after the 2nd stroke.
Let $a_4$ be the amount of air after the 3rd stroke.

Step 2: Calculating the sequence of air amounts
The pump removes $\frac{1}{4}$ of the air present in the cylinder at each step.

Initial amount: $a_1 = V$

After the 1st stroke ($a_2$):
$a_2 = V - \frac{1}{4}V = \frac{3}{4}V$

After the 2nd stroke ($a_3$):
The pump removes $\frac{1}{4}$ of the air remaining, which is $a_2$.
$a_3 = a_2 - \frac{1}{4}a_2 = \frac{3}{4}a_2$
Substituting $a_2 = \frac{3}{4}V$:
$a_3 = \frac{3}{4} \times (\frac{3}{4}V) = \frac{9}{16}V$

After the 3rd stroke ($a_4$):
$a_4 = a_3 - \frac{1}{4}a_3 = \frac{3}{4}a_3$
Substituting $a_3 = \frac{9}{16}V$:
$a_4 = \frac{3}{4} \times (\frac{9}{16}V) = \frac{27}{64}V$

Step 3: Checking for Arithmetic Progression
A sequence is an Arithmetic Progression if the difference between consecutive terms is constant (i.e., $a_{n+1} - a_n = d$, where $d$ is the common difference).

Calculate the first difference ($d_1$):
$d_1 = a_2 - a_1 = \frac{3}{4}V - V = -\frac{1}{4}V$

Calculate the second difference ($d_2$):
$d_2 = a_3 - a_2 = \frac{9}{16}V - \frac{3}{4}V$
To subtract, find a common denominator (16):
$d_2 = \frac{9}{16}V - \frac{12}{16}V = -\frac{3}{16}V$

Step 4: Comparison and Conclusion
Since $d_1 \neq d_2$ (because $-\frac{1}{4}V \neq -\frac{3}{16}V$), the difference between consecutive terms is not constant.

[Definition of an Arithmetic Progression: A sequence of numbers is an AP if the difference between any two consecutive terms is constant.]

Final Answer: The list of numbers does not form an Arithmetic Progression because the difference between consecutive terms is not constant.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.1


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


Chapters in CBSE - Class 10 Mathematics


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