Recurring deposit accounts

A Recurring Deposit (RD) account is a banking scheme where a saver deposits a fixed sum every month for a fixed period, and the bank pays interest on the whole set of instalments when the account matures. It sounds like ordinary compound interest, but it is not: every instalment sits in the bank for a different length of time, so the interest calculation has to add up many small periods rather than apply one growth factor. This is the banking topic inside the Commercial Mathematics unit of ICSE Class 10 Mathematics. The syllabus expects you to compute the interest earned on an RD account and the maturity value received at the end, using a dedicated RD formula — not the compound interest formula of Class 9, which never reappears here. Fixed deposits, savings-account slabs and EMI mathematics are outside this syllabus; only the RD calculation is examinable, and it is a reliably scoring topic once the formula is understood rather than memorised.

The core insight

Because each monthly instalment earns interest for a different number of months, total RD interest is found by summing instalment-months, $I = P\cdot\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100}$ , and the maturity value adds back the total deposited, $MV = Pn + I$ .

The core idea

In an RD account a saver deposits a fixed sum $P$ — the monthly instalment — at the start of every month, for $n$ months, at an annual interest rate of $r\%$ . The bank does not treat all instalments the same way. The first instalment sits in the account for close to the full $n$ months and earns interest on all of it; the last instalment, deposited just before the account closes, earns interest for only about one month. The total interest is therefore a sum over all these different durations, not a single compound-growth calculation. That sum collapses into a compact formula:

I = P\cdot\frac{n(n+1)}{2\times12}\times\frac{r}{100}

where:

$P$ = monthly instalment (the fixed amount deposited each month)
$n$ = number of months the account runs
$r$ = annual rate of interest, as a percentage
$I$ = total interest earned over the whole period

The fraction $\dfrac{n(n+1)}{2}$ is the sum of the first $n$ natural numbers — it counts the total number of "instalment-months" of money sitting in the bank. Dividing by 12 converts the annual rate $r\%$ into a monthly rate, since interest accrues month by month, not year by year. Once the interest is known, the maturity value — the total amount handed back to the depositor when the account closes — is simply everything deposited plus everything earned:

MV = P\times n + I

where: $MV$ is the maturity value, $P\times n$ is the total principal deposited across all $n$ months, and $I$ is the interest from the first formula. Missing this addition is one of the most common scoring errors in this topic: interest alone is not the maturity value.

Figure

A staircase diagram showing each monthly instalment held in the bank for a different number of months, from n months down to 1 month

Derivation

Derivation ABuilding the RD interest formula from instalment-months

The goal is to derive $I = P\cdot\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100}$ from first principles, using nothing beyond simple interest reasoning and the sum of the first $n$ natural numbers, so that the formula can be reconstructed under exam pressure instead of half-remembered and confused with compound interest.

Step 1. Track how long each instalment stays in the bank. By the usual RD convention, the first instalment (deposited at the start of month 1) earns interest for $n$ months, the second instalment for $(n-1)$ months, the third for $(n-2)$ months, and so on, down to the very last instalment, deposited one month before maturity, which earns interest for just 1 month.

Step 2. Apply the simple-interest formula to a single instalment held for $t$ months. Simple interest on a sum $P$ at annual rate $r\%$ for $t$ months is $P\times\dfrac{r}{100}\times\dfrac{t}{12}$ , since $t/12$ converts months into a fraction of a year.

Step 3. Add up the interest from every instalment. The total interest is the sum over $t = n, (n-1), (n-2), \ldots, 1$ of $P\times\dfrac{r}{100}\times\dfrac{t}{12}$ . Since $P$ , $r$ and $12$ are the same for every term, they can be pulled out of the sum, leaving $I = P\times\dfrac{r}{1200}\times\left[n + (n-1) + (n-2) + \cdots + 1\right]$ .

Step 4. Evaluate the bracketed sum. The sum of the first $n$ natural numbers is the well-known result $n + (n-1) + \cdots + 1 = \dfrac{n(n+1)}{2}$ . Substituting this back gives $I = P\times\dfrac{r}{1200}\times\dfrac{n(n+1)}{2} = P\cdot\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100}$ , which is exactly the RD interest formula. The formula is therefore not an arbitrary rule but a direct consequence of adding up $n$ separate simple-interest calculations, one for each instalment.

Worked example

Rekha deposits ₹500 every month in a Recurring Deposit account for 2 years (24 months), at an annual interest rate of 8%. Find the interest earned and the maturity value.

Step 1. Identify the values from the problem: $P = 500$ , $n = 24$ , $r = 8$ .

Step 2. Apply the interest formula: $I = P\cdot\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100} = 500\times\dfrac{24\times25}{24}\times\dfrac{8}{100}$ .

Step 3. Simplify. $\dfrac{24\times25}{2\times12} = \dfrac{600}{24} = 25$ , so $I = 500\times25\times\dfrac{8}{100} = 12500\times0.08 = ₹1{,}000$ .

Step 4. Find the maturity value: $MV = P\times n + I = 500\times24 + 1000 = 12000 + 1000 = ₹13{,}000$ . Rekha earns ₹1,000 in interest and receives a maturity value of ₹13,000.

Competitive stretch

Suresh opened an RD account for 3 years (36 months) at 6% per annum and received ₹39,330 at maturity. Find his monthly instalment and the interest earned.

Step 1. Write the maturity value with $P$ factored out: $MV = P\left[n + \dfrac{n(n+1)}{24}\times\dfrac{r}{100}\right]$ .

Step 2. Substitute $n = 36$ , $r = 6$ : $\dfrac{36\times37}{24}\times\dfrac{6}{100} = 55.5\times0.06 = 3.33$ , so the bracket is $36 + 3.33 = 39.33$ .

Step 3. Solve for $P$ : $39{,}330 = P\times39.33 \Rightarrow P = \dfrac{39{,}330}{39.33} = ₹1{,}000$ .

Step 4. Interest = maturity value minus total deposited = $39{,}330 - (1000\times36) = 39{,}330 - 36{,}000 = ₹3{,}330$ . Suresh's monthly instalment was ₹1,000, and he earned ₹3,330 in interest. Factoring $P$ out of both terms of the $MV$ formula, so that $MV$ becomes $P$ multiplied by one combined number, is the standard technique whenever the instalment itself is unknown.

Common traps

Using the compound interest formula $A = P(1+r/100)^n$ for RD accounts — the RD formula has a completely different, instalment-based structure.
Forgetting the $2\times12$ in the denominator and dividing by 2 alone — this mixes up the annual rate with the monthly compounding period and gives a wrong, too-large interest.
Confusing $n$ (number of months) with the number of years given in the question — always convert the deposit period to months first.
Reporting the interest $I$ as the final answer when the question asks for maturity value — maturity value is principal plus interest, $MV = Pn + I$ , not $I$ alone.

Ask the tutor: if the numbers in an RD problem aren't matching up cleanly, ask the AI tutor to re-derive the interest step by step with your specific values.

Why is the RD formula different from the simple interest formula I already know?

It is built from simple interest, just applied many times over. Each monthly instalment sits in the bank for a different length of time, so you cannot use one $P$ , one $t$ and one simple-interest calculation the way you would for a single lump-sum deposit. Instead you add up $n$ separate simple-interest amounts, one per instalment, and that sum simplifies to $\dfrac{n(n+1)}{2}$ instalment-months, which is exactly where the RD formula's numerator comes from.

Do I need to remember whether the first instalment earns interest for n months or (n-1) months?

For ICSE Class 10 exam purposes, always use the standard formula $I = P\cdot\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100}$ exactly as given — it already assumes the conventional counting (first instalment for close to $n$ months, last for 1 month) baked into the derivation. You do not need to re-derive it from scratch in an exam; just substitute $P$ , $n$ and $r$ correctly and the formula handles the counting for you.

What if a question gives the rate of interest as unknown instead of P or n?

The same technique applies: write down the interest or maturity value formula with every known value substituted, leaving $r$ as the only unknown, then solve the resulting equation for $r$ . Since $r$ appears as a simple multiplying factor $r/100$ in the formula, isolating it is usually a one-step division once everything else is substituted, exactly as $P$ was isolated in the stretch example above.