In which of the following situations, do the lists of numbers involved form an $ A P $ ? Give reasons for your answers.
The fee charged from a student every month by a school for the whole session, when the monthly fee is Rs 400.

To do:

We have to check whether in the given situations the list of numbers involved forms an \( A P \).

Solution:

(i) The fee charged from a student every month by a school for the whole session, when the monthly fee is Rs 400.

This implies, the sequence formed(in Rs.) is $400, 400, 400, .......$

In the given sequence,

$a_1=400, a_2=400, a_3=400$

$a_2-a_1=400-400=0$ 

$a_3-a_2=400-400=0$

Here,

$a_3-a_2=a_2-a_1$

Therefore,

The list of numbers involved in the given situation forms an \( A P \).

(ii) The fee charged every month by a school from Classes I to XII, when the monthly fee for Class I is Rs 250 , and it increases by Rs 50 for the next higher class.

This implies, the sequence formed(in Rs.) is $250, (250+50), (250+50+50), .......$

The sequence is $250, 300, 350, .......$

In the given sequence,

$a_1=250, a_2=300, a_3=350$

$a_2-a_1=300-250=50$ 

$a_3-a_2=350-300=50$

Here,

$a_3-a_2=a_2-a_1$

Therefore,

The list of numbers involved in the given situation forms an \( A P \).

(iii) We know that,

Simple interest $=\frac{\text { Principal } \times \text { Rate } \times \text { Time }}{100}$

$=\frac{1000 \times 10 \times 1}{100}$

$=100$

This implies,

The amount of money in the account of Varun at the end of each year(in Rs.) is,

$1000, (1000+100 \times 1),(1000+100 \times 2),(1000+100 \times 3), \ldots $

The sequence formed is,

$1000,1100,1200,1300, \ldots $

In the given sequence,

$a_1=1000, a_2=1100, a_3=1200$

$a_2-a_1=1100-1000=100$ 

$a_3-a_2=1200-1100=100$

Here,

$a_3-a_2=a_2-a_1$

Therefore,

The list of numbers involved in the given situation forms an \( A P \).

(iv) Let the number of bacteria in a certain food be $x$.

Given that the bacteria double in every second.

This implies,

The number of bacteria after every second is,

$x, 2 x, 2(2 x), 2(2 \times 2 x), \ldots$

The sequence formed is,

$x, 2 x, 4 x, 8 x, \ldots$

In the given sequence,

$a_1=x, a_2=2x, a_3=4x, a_4=8x$

$a_2-a_1=2x-x=x$ 

$a_3-a_2=4x-2x=2x$

$a_4-a_3=8x-4x=4x$

Here,

$a_3-a_2≠a_2-a_1$

Therefore,

The list of numbers involved in the given situation does not form an \( A P \).