#### i)In which of the following situations, do the lists of numbers involved form an AP? 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.ii)In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.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.iii)In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.The amount of money in the account of Varun at the end of every year when Rs 1000 is deposited at simple interest of 10% per annum.iv)In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.The number of bacteria in a certain food item after each second, when they double in every second.

i)

Solution.

According to the question:

It is clear that the monthly fee charged from students is = 400 Rs.

Hence series is ; 400, 400, 400, 400 ……

the difference of the successive terms is same so It forms an AP with common difference d = 400 – 400 = 0

ii)

Solution.

According to the question:

Fees started Rs. 250 and increased by 50 Rs. from I to XII is

$\\Here first term( a\textsubscript{1})=250 common difference (d)=50 \\ a\textsubscript{1}=250\\ a\textsubscript{2}=250+50\\ =300\\ a\textsubscript{3}=300+50\\ =350\\ a\textsubscript{4}=350+50\\ =400\\ So the A.P. 250, 300, 350, 400 \ldots \\$

It forms an AP with common difference 50.

iii)

Solution.         We know that: Simple interest = (P × R × T)/100

$\text{Here } P = 1000 \text{ Rs, } R = 10\% \text{ p.a., } T = 1 \text{ year} \\ \text{SI} = \frac{1000 \times 10 \times 1}{100} = 100 \text{ Rs} \\ \text{Here common difference } (d) = 100 \text{ Rs and first term } (a_1) = 1000 \text{ Rs} \\ a_1 = 1000 \text{ Rs} \\ a_2 = 1000 + 100 = 1100 \text{ Rs} \\ a_3 = 1000 + 200 = 1200 \text{ Rs} \\ a_4 = 1000 + 300 = 1300 \text{ Rs} \\ \text{So the AP is: } 1000, 1100, 1200, 1300, \ldots$

This is definite form an AP with common difference of SI=100 Rs.

iv)

Solution.

Let the number of bacteria = x

According to the question:

They double in every second.

$\text{Here } a_1 = x, \quad a_2 = 2x, \quad a_3 = 4x, \quad a_4 = 8x, \ldots \\ x, 2x, 4x, 8x, \ldots$

Since the difference between the successive terms is not same therefore it does not form an AP.