Banking: Recurring Deposit Accounts — chapter notes

In a recurring deposit (RD) account a depositor pays a fixed monthly instalment $P$ for a fixed number of months $n$. The bank pays simple interest at $r\%$ per annum on each instalment for the time it stays in the account, and returns the instalments with the interest at maturity.

• The 1st instalment earns interest for $n$ months, the 2nd for $n - 1$ months, …, the last for $1$ month.
• A recurring deposit suits people who can save a little every month rather than a lump sum.

At the end of month $k$ the balance is $kP$. The sum of the balances,

earns simple interest for one month. It is called the principal for one month. Example: ₹$500$ a month for $6$ months: $500 \times 21 = 10500$.

Simple interest on the principal for one month, for $\dfrac{1}{12}$ of a year at $r\%$:

• $n$ is in months (1.5 years $= 18$, 2 years $= 24$, 3 years $= 36$); $r$ is per annum.
• Example: ₹$800$ a month for $18$ months at $7\%$: $I = 800 \times \dfrac{18 \times 19}{24} \times \dfrac{7}{100} = 798$.
• Keep paise: ₹$1250$ for $24$ months at $6.5\%$ gives ₹2,031.25.

With $P$ and $r$ fixed, the interest is proportional to $n(n+1)$, not to $n$. Doubling the period from $12$ to $24$ months multiplies $n(n+1)$ from $156$ to $600$: almost four times.

For the same total deposited, a longer deposit with smaller instalments earns more: ₹$600$ for $24$ months earns ₹$1200$ at $8\%$, ₹$1200$ for $12$ months only ₹$624$.

The maturity value is the total deposited plus the interest:

Example: ₹$1500$ a month for $3$ years at $7.5\%$: deposits $54000$, interest $6243.75$, maturity value ₹60,243.75. If the instalment is doubled (same $n$ and $r$), the maturity value doubles.

• Instalment: write the interest as a multiple of $P$ (e.g. $n = 24$, $r = 9$: $I = 2.25P$), so $\text{MV} = 26.25P$; divide.
• Rate: first $I = \text{MV} - Pn$; then write $I$ as a multiple of $r$ (e.g. $P = 2000$, $n = 15$: $I = 200r$); divide.
• Answers often have paise or decimals (a rate of $6.5\%$); give them exactly.
• If an instalment must reach at least a target, round up to the next rupee.

With $n$ unknown:

• Interest given: $n(n+1) = \dfrac{2400I}{Pr}$; find two consecutive whole numbers with this product.
• Maturity value given: $\text{MV} = Pn + \dfrac{Pr}{2400}\,n(n+1)$, a quadratic in $n$. Divide by the coefficient of $n^2$, factorise (or use the quadratic formula) and reject the negative root.

Example: ₹$1200$ a month at $8\%$ maturing to ₹18,960: $n^2 + 301n - 4740 = 0$, $(n - 15)(n + 316) = 0$, so $n = 15$.

Always check by working forwards.