Why Saving ₹1000 a Month Earns Less Interest Than Depositing ₹12000 at Once

In a recurring deposit (RD) account, you deposit the same amount every month for a fixed number of months. At the end, the bank returns all your deposits plus simple interest.

Each instalment earns interest only for the months it stays in the bank. For a deposit of ₹$P$ a month for $n$ months, the first instalment stays $n$ months, the second $n-1$ months, and the last only $1$ month.

This chapter covers the interest formula, maturity value, and finding the instalment, rate or time.

**Use $I=P\times \frac{n(n+1)}{2\times 12}\times \frac{r}{100}$, where $P$ is the monthly instalment, $n$ the number of months and $r$ the yearly rate of interest.

Where the formula comes from.** Adding the months each instalment stays:

$$n+(n-1)+\cdots +2+1=\frac{n(n+1)}{2}$$

So the deposits earn as much as ₹$P$ kept for $\frac{n(n+1)}{2}$ months, which is $\frac{n(n+1)}{2\times 12}$ years.

Worked example. Riya deposits ₹800 a month for $2$ years at $6\%$ per annum.

$$n=24\frac{n(n+1)}{2\times 12}=\frac{24\times 25}{24}=25$$
$$I=800\times 25\times \frac{6}{100}=₹1200$$

An everyday example. Many families open an RD at a bank or post office to save a fixed sum each month for a school trip or a festival.

The substance. **$\frac{n(n+1)}{2}$ is the sum of the first $n$ natural numbers**, so the formula is simple interest applied to a growing pile of monthly deposits.

**The maturity value is the total of all deposits plus the interest: $MV=P\times n+I$.

Worked example 1.** Continuing Riya's account:

$$\text{total deposits}=800\times 24=₹19200MV=19200+1200=₹20400$$

Worked example 2. Arjun deposits ₹1500 a month for $3$ years at $8\%$ per annum.

$$n=36\frac{n(n+1)}{2\times 12}=\frac{36\times 37}{24}=55.5$$
$$I=1500\times 55.5\times \frac{8}{100}=₹6660$$
$$MV=1500\times 36+6660=54000+6660=₹60660$$

An everyday example. Parents saving for college admission fees often choose an RD period that ends just before the fees are due, so the maturity value is ready in time.

The misconception. The maturity value is not just the interest. Students sometimes stop at $I$; the question almost always wants the deposits added back.

**Write the maturity value in terms of $P$, collect the terms containing $P$, and divide.**

$$MV=Pn+P\times \frac{n(n+1)}{2\times 12}\times \frac{r}{100}=P\left(n+\frac{n(n+1)}{24}\times \frac{r}{100}\right)$$

Worked example. An RD account for $1$ year at $8\%$ per annum has a maturity value of ₹7512. Find the monthly instalment.

$$n=12\frac{12\times 13}{24}=6.56.5\times \frac{8}{100}=0.52$$
$$7512=P(12+0.52)=12.52PP=\frac{7512}{12.52}=₹600$$

Check: $I=600\times 6.5\times \frac{8}{100}=₹312$, and $600\times 12+312=₹7512$.

An everyday example. A student wanting ₹7512 in a year for a new bicycle can use this working to find how much to deposit each month.

The substance. **Factorising $P$ out first** turns the problem into one simple division.

**Substitute everything known into the interest or maturity formula and solve for $r$ as a linear equation, or for $n$, which usually leads to a quadratic.

Worked example 1 — rate.** Meena deposits ₹500 a month for $1$ year and earns ₹195 interest. Find the rate.

$$195=500\times \frac{12\times 13}{24}\times \frac{r}{100}=500\times 6.5\times \frac{r}{100}=32.5rr=\frac{195}{32.5}=6\%$$

Worked example 2 — months, from interest. Kabir deposits ₹400 a month at $10\%$ and earns ₹1000 interest. Find $n$.

$$1000=400\times \frac{n(n+1)}{24}\times \frac{10}{100}=\frac{5}{3}n(n+1)n(n+1)=600=24\times 25n=24$$

Worked example 3 — months, from maturity value. The same account matures at ₹10600.

$$400n+\frac{5}{3}n(n+1)=10600\Rightarrow n^2+241n-6360=0\Rightarrow (n-24)(n+265)=0$$

**Since $n$ cannot be negative, $n=24$ months.

An everyday example. Comparing RD schemes from two banks, you can use the interest earned to work out which offers the better rate.

The substance. Reject the negative root** — a number of months must be a positive whole number.