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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 \( A P \).