What is banking?

• The business of receiving money from depositors (or account holders), safeguarding, and lending money to businesses or individuals is called banking.
• Therefore Banks are institutions that carry out the business of taking deposits and lending money.
• When people deposit money, based on the scheme under which they deposit money, they get a return on their money.
• Similarly, when the bank lends money to people or businesses, banks charge a rate of interest on the amount of money given to us.

The difference between the two is what the bank earns adjusted to their operational costs. In a very simple way we could say:

Banks Earnings = Earning on Money Invested – Interest paid to account holders  –  Operational cost

Off-course, banks have to take a banking license to start bank operations.

Types of Accounts: There are many types of accounts but from a course perspective, we will look at the following accounts:

1. Savings Bank Account
2. Recurring Deposit Accounts

Saving Bank Account

A person can deposit and withdraw money at will. The person gets a certain interest on the deposits, which could change with a change in market conditions.

How do we calculate the interest on the deposit?

Nowadays, because of the powerful computers, banks are able to calculate interest on a day-to-day basis. However, for our syllabus, we would calculate interest on a monthly basis. The concept is the same though. Here is how we will do it.

1. Find the minimum balance on the $\displaystyle 10^{th}$ day and up to the last day of each month. This minimum balance becomes the principal of the month.
2. Add all such Principal amounts obtained for different months of a particular period in consideration.
3. Now calculate the simple interest on the Principal obtained in Step 2 for one month at the prevailing rate of interest at that time.

Recurring Deposit Account

In this type of deposit, the account holder deposits a specified amount in the account every month for a fixed period of time. It could be three months to say 10 years. The time period is decided by the bank.

At the expiry of the period, the person gets a lump sum of money which includes the money that was deposited and the interest (compounded quarterly) that the money has earned over a period of time.

The formula that we use for calculating the maturity value of the recurring deposit is:

Maturity Amount  =  Total Sum Deposited  +  Interest Earned

If $\displaystyle P$ is deposited every month in the bank for $\displaystyle n$ months and $\displaystyle r\%$ is the rate of interest per year, then

$\displaystyle I=P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

Total Sum Deposited $\displaystyle = P \times n$

$\displaystyle \text{Maturity Value }= P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

Proof: If $\displaystyle P$ is deposited every month in the bank for $\displaystyle n$ months and $\displaystyle r\%$ is the rate of interest per year, then

Maturity amount for $\displaystyle P$ deposited in the $\displaystyle 1^{st}$ month after $\displaystyle n$ months

$\displaystyle = P + P \times \frac{r}{100} \times \frac{n}{12}$

Maturity amount for $\displaystyle P$ deposited in the $\displaystyle 2^{nd}$ month after $\displaystyle (n-1)$ months

$\displaystyle = P + P \times \frac{r}{100} \times \frac{(n-1)}{12}$

Maturity amount for $\displaystyle P$ deposited in the $\displaystyle 3^{rd}$ month after $\displaystyle (n-1)$ months

$\displaystyle = P + P \times \frac{r}{100} \times \frac{(n-2)}{12}$

$\displaystyle \cdots$

Maturity amount for $\displaystyle P$ deposited in the $\displaystyle (n-1)^{th}$ month after $\displaystyle (n-1)$ months

$\displaystyle = P + P \times \frac{r}{100} \times \frac{2}{12}$

Maturity amount for $\displaystyle P$ deposited in the $\displaystyle (n)^{th}$ month after $\displaystyle (n-1)$ months

$\displaystyle = P + P \times \frac{r}{100} \times \frac{1}{12}$

Therefore

$\displaystyle \text{Maturity amount } = P \times n + P \frac{r}{100} \Big( \frac{n}{12} + \frac{n-1}{12} + \frac{n-2}{12} + \cdots + \frac{2}{12} + \frac{1}{12} \Big)$

$\displaystyle = P \times n + P \frac{r}{100 \times 12 } \Big( n + (n-1) + (n-2) + \cdots + 2 + 1 \Big)$

$\displaystyle = P \times n + P \frac{n(n+1)}{2 \times 12} \frac{r}{100}$