Question

In which of the following situations, the sequence formed will form an A.P.?
A. Number of students left in the school auditorium from the total strength of $1000$ students when they leave the auditorium in batches of $25$ .
B. The amount of money in the account every year when $\text{Rs}\text{. 100}$ is deposited annually to accumulate compound interest at $4\%$ per annum.

Answer 1

We will apply the condition given in the question as for A we will keep subtracting $25$ so that we get a particular series then we will see if there is a common difference then that series is in A.P. , and for the second option that is B and we will apply the formula for calculating the amount that is $A=P{{\left( 1+\left( \dfrac{R}{100} \right) \right)}^{n}}$ and get the series and keep varying the value of $n$ and for this series as well we will look for a common difference and check whether it is an AP or not.

Complete step-by-step answer:
We will start with considering our first case
A. It is given that the total strength of the students in the auditorium = $1000$ and it is given that the students leave in a group of $25$ students.
Now, the number of students left in the auditorium when first batch of 25 students leaves the auditorium: $1000-25=975$
After that, the number of students left in the auditorium when second batch of 25 students leaves the auditorium: $975-25=950$
Finally, the number of students left in the auditorium when the third batch of 25 students leaves the auditorium: $950-25=925$ and so on........
Thus, the number of students left in auditorium at different stages are: $1000,975,950,925........$ Clearly, it is an arithmetic progression with first term as $\left( a \right)$ that is $1000$ and the common difference equals to $-25$ .

Now, we will see the B part:
B. We know that, if $P$ is the principal and $R\%$ per annum is the rate of interest compounded annually then, the amount $A$ at the end of $n$ years: $A=P{{\left( 1+\left( \dfrac{R}{100} \right) \right)}^{n}}$
Now, it is given that $P=100,R=4\%$,
Therefore, $A=100{{\left( 1+\left( \dfrac{4}{100} \right) \right)}^{n}}\Rightarrow A=100{{\left( 1+0.04 \right)}^{n}}$
Now, the amount of money in the account at the end of different years is given by:
$\begin{aligned} & 100\left( 1+0.04 \right),100{{\left( 1+0.04 \right)}^{2}},100{{\left( 1+0.04 \right)}^{3}},...... \\\ & or \\\ & 104,108.16,112.48........ \\\ \end{aligned}$
for $n=1,2,3.....$
As we can see that this is not forming an arithmetic progression.

Hence, only A is in A.P.

Note: Always describe the steps you perform in these types of questions as when you are applying the conditional formula asked in question. Also, if that particular sequence turns out that it is in arithmetic progression then always write the first term and common difference at the end to conclude your answer. Try to be careful while calculating amount in the B part student can make mistakes when $n>1$ as we will have to take squares or cubes of the number obtained from $A=P{{\left( 1+\left( \dfrac{R}{100} \right) \right)}^{n}}$