MathsICSE Class 10 · Banking

Finding the Monthly Instalment and Time of a Recurring Deposit (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

To find the monthly instalment P, put the known values in I = P × n(n + 1)/24 × r/100 (or MV = Pn + I) and solve the linear equation for P. To find the time, the same formula gives a quadratic in n: n(n + 1) = 2400I/(Pr). Solve it and keep the positive whole number of months.

Finding P is a linear equation, but finding n gives n(n+1) = k: find two consecutive numbers, reject the negative root, and check the units.

Illustration drawn for QuestionBuddy.

Understand Monthly Instalment and Time of a Recurring Deposit

The recurring deposit formula links five quantities: the monthly instalment $P$ , the number of months $n$ , the rate $r %$ per annum, the interest $I$ and the maturity value $M V$ :

$I = P \times \frac{n ( n + 1 )}{2 \times 12} \times \frac{r}{100}, M V = P n + I .$

Besides the rate, a board question may hide the instalment or the time.

**Finding the instalment $P$ .** With $n$ and $r$ known, $\frac{n ( n + 1 )}{24} \times \frac{r}{100}$ is just a number, so the interest equation is linear in $P$ :

$P = \frac{I}{\frac{n ( n + 1 )}{24} \times \frac{r}{100}} .$

If the maturity value is given instead of the interest, both terms contain $P$ :

$M V = P [n + \frac{n ( n + 1 )}{24} \times \frac{r}{100}] \Rightarrow P = \frac{M V}{n + \frac{n ( n + 1 )}{24} \times \frac{r}{100}} .$

For 2 years at 8% the bracket is $24 + 25 \times 0.08 = 26$ , so every rupee of instalment grows to ₹26 by maturity.

**Finding the time $n$ .** Now $n$ appears as $n (n + 1)$ , so the equation is a quadratic. When the interest is given:

$n (n + 1) = \frac{2400 I}{P r} .$

Solve $n^{2} + n - k = 0$ by factorising: you need two consecutive numbers whose product is $k$ . For $k = 156$ , $12 \times 13 = 156$ , so $n = 12$ . Reject the negative root, since time cannot be negative. This is exactly what 2024 Q4(i) asked; the examiners reported errors in forming and solving $n^{2} + n - 156 = 0$ and answers left as '12 months' when the question asked for years.

When the maturity value is given, the quadratic also has the $P n$ term:

$P n + P \times \frac{n ( n + 1 )}{24} \times \frac{r}{100} = M V,$

which, after multiplying out and dividing by a common factor, gives $n^{2} + bn - c = 0$ with a larger middle coefficient. Factorise by looking for factors of $c$ that differ by $b$ , or use the quadratic formula.

Answer in the form asked. If the question asks for the time in years, divide the months by 12.

Monthly Instalment and Time of a Recurring Deposit: formulas to remember

Instalment from the interest: $P = \frac{I}{\frac{n ( n + 1 )}{24} \times \frac{r}{100}}$; Linear in $P$ .
Instalment from the maturity value: $P = \frac{M V}{n + \frac{n ( n + 1 )}{24} \times \frac{r}{100}}$; The bracket is what ₹1 a month grows to.
Time from the interest: $n (n + 1) = \frac{2400 I}{P r}$; Find two consecutive whole numbers with this product.
Time from the maturity value: $P n + P \times \frac{n ( n + 1 )}{24} \times \frac{r}{100} = M V$; Rearrange to $n^{2} + bn - c = 0$ .
Quadratic formula (when factorising is hard): $n = \frac{- b + b ^{2} + 4 c}{2}$; For $n^{2} + bn - c = 0$ ; take the positive root.

How to solve Monthly Instalment and Time of a Recurring Deposit questions

Write the recurring deposit formula, substitute everything you know, and solve for the one unknown.

1List the known quantities; convert the time to months if it is known.
2If the maturity value is given and the instalment or time is unknown, do not subtract yet: write $M V = P n + I$ in full.
3For $P$ : collect the terms in $P$ and divide.
4For $n$ from the interest: write $n (n + 1) = \frac{2400 I}{P r}$ and find consecutive factors.
5For $n$ from the maturity value: multiply out, simplify to $n^{2} + bn - c = 0$ , and factorise or use the formula.
6Reject negative or fractional values of $n$ ; give the time in the unit asked.
7Check by putting your answer back into the formula.

Monthly Instalment and Time of a Recurring Deposit in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Choosing the instalment for a goal

Suppose you want exactly ₹6,195 in one year to buy a bicycle, and your bank offers a recurring deposit at 6% per annum.

How much must you deposit each month?

Try it first, then show the solutionSolution

$n = 12$ : $\frac{12 \times 13}{24} \times \frac{6}{100} = 6.5 \times 0.06 = 0.39$ .
$M V = P (12 + 0.39) = 12.39 P$ .
$12.39 P = 6195 \Rightarrow P = 500$ .

Answer: ₹500 a month.

Memory hook: Work out what ₹1 a month becomes, then divide your goal by it.

Finding the time from the interest

Imagine you deposit ₹1,000 a month in a recurring deposit at 9% per annum and the interest at maturity is ₹1,425.

For how many months was the account held?

Try it first, then show the solutionSolution

$n (n + 1) = \frac{2400 \times 1425}{1000 \times 9} = 380$ .
$19 \times 20 = 380$ , so $n = 19$ (rejecting $n = - 20$ ).

Answer: 19 months.

Memory hook: n(n + 1) is a product of neighbours: hunt for two consecutive numbers.

Tricks to remember Monthly Instalment and Time of a Recurring Deposit

Short memory aids. Say each one aloud once, then use it on a question the same day.

Neighbours multiply
$n (n + 1)$ is the product of two neighbours. Take the square root of the right side and try the whole numbers either side: $380 \approx 19.5$ , so try $19 \times 20$ .
Use it when: Solving for $n$ quickly.
₹1 grows to …
For 2 years at 8%, ₹1 a month grows to ₹26 at maturity. Goal ÷ growth factor = instalment.
Use it when: Instalment from maturity value.
Answer in the unit asked
Circle 'years' or 'months' in the question and convert at the end.
Use it when: Avoiding the 2024 Q4(i) slip of leaving 12 months when years were asked.

Monthly Instalment and Time of a Recurring Deposit: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Solving n(n + 1) = 600 · 1 mark · MCQBoard level

In a recurring deposit question, the time n in months satisfies $n (n + 1) = 600$ . The value of n is:
(a) 20
(b) 24
(c) 25
(d) 30

Try it first, then show the solutionSolution

$600 \approx 24.5$ ; try $24 \times 25 = 600$ .
So $n = 24$ .

Answer: (b) 24

Example 2: Instalment from the interest · 3 marks · Short answerBoard level

A recurring deposit for 2 years at 8% per annum earns ₹1,600 as interest. Find the monthly instalment and the maturity value.

Try it first, then show the solutionSolution

$I = P \times \frac{24 \times 25}{24} \times \frac{8}{100} = P \times 25 \times 0.08 = 2 P$ .
$2 P = 1600 \Rightarrow P = 800$ .
$M V = 800 \times 24 + 1600 = 19200 + 1600 = 20800$ .

Answer: ₹800 a month; maturity value ₹20,800.

Example 3: Instalment from the maturity value · 3 marks · Short answerHarder board level

A recurring deposit for 1 year at 9% per annum matures to ₹12,585. Find the monthly instalment.

Try it first, then show the solutionSolution

$I = P \times 6.5 \times \frac{9}{100} = 0.585 P$ .
$M V = 12 P + 0.585 P = 12.585 P$ .
$12.585 P = 12585 \Rightarrow P = 1000$ .

Answer: ₹1,000 a month.

Example 4: Time from the interest · 4 marks · Short answerHarder board level

Mohan deposits ₹1,500 per month in a recurring deposit at 8% per annum and receives ₹1,360 as interest at maturity. Find the time for which the account was held.

Try it first, then show the solutionSolution

$1360 = 1500 \times \frac{n ( n + 1 )}{24} \times \frac{8}{100} = 5 n (n + 1)$ .
$n (n + 1) = 272 \Rightarrow n^{2} + n - 272 = 0$ .
$(n - 16) (n + 17) = 0 \Rightarrow n = 16$ (time cannot be negative).
Check: $1500 \times \frac{16 \times 17}{24} \times 0.08 = 1360$ .

Answer: 16 months (1 year 4 months).

Example 5: Time from the maturity value · 4 marks · Short answerHOTS / competency

A man deposits ₹600 per month in a recurring deposit at 10% per annum and receives ₹11,655 at maturity. Find the number of months.

Try it first, then show the solutionSolution

$11655 = 600 n + 600 \times \frac{n ( n + 1 )}{24} \times \frac{10}{100} = 600 n + 2.5 n (n + 1)$ .
$2.5 n^{2} + 602.5 n - 11655 = 0$ ; multiply by 2 and divide by 5: $n^{2} + 241 n - 4662 = 0$ .
$4662 = 18 \times 259$ and $259 - 18 = 241$ , so $(n - 18) (n + 259) = 0$ .
$n = 18$ . Check: $10800 + 600 \times 14.25 \times 0.1 = 10800 + 855 = 11655$ .

Answer: 18 months (1½ years).

Example 6: Case study: saving for a target · 4 marks · Case-basedHOTS / competency

Priya wants ₹52,000 at the end of 2 years from a recurring deposit at 8% per annum.
(a) Find the monthly instalment she needs.
(b) If she can save only ₹1,500 a month, what will she receive?
(c) What percentage of her deposits does she earn as interest in (a)?

Try it first, then show the solutionSolution

Growth factor $= 24 + 25 \times 0.08 = 26$ .
(a) $26 P = 52000 \Rightarrow P = 2000$ .
(b) $1500 \times 26 = 39000$ .
(c) Deposits $= 48000$ , interest $= 4000$ ; $\frac{4000}{48000} \times 100 = 8 \frac{1}{3} %$ .

Answer: (a) ₹2,000 (b) ₹39,000 (c) $8 \frac{1}{3} %$ of her deposits.

Where students lose marks in Monthly Instalment and Time of a Recurring Deposit

The slips examiners see most often, and what to write instead.

Mistake: Subtracting a guess for P × n before P is known
When $P$ (or $n$ ) is unknown, $P n$ is unknown too.
Wrong working
$I = 12585 - 12000 = 585$ (assuming $P = 1000$ ).
Right working
$12 P + 0.585 P = 12585 \Rightarrow P = 1000$ .
Do this: Write $M V = P n + I$ with the unknown in both terms.
Usually costs: The method marks.
Mistake: Errors in forming or solving the quadratic
Arithmetic slips with 2400 and the rate are common.
Wrong working
$1040 = 2000 \times \frac{n ( n + 1 )}{24} \times 8 \Rightarrow n (n + 1) = 1.56$ .
Right working
$1040 = 2000 \times \frac{n ( n + 1 )}{24} \times \frac{8}{100} \Rightarrow n (n + 1) = 156$ .
Do this: Keep the $\frac{r}{100}$ and simplify the constant carefully.
Usually costs: The examiners named this for 2024 Q4(i).
Mistake: Answering in months when years are asked
The formula gives $n$ in months.
Wrong working
Time = 12 months. (Question asked for years.)
Right working
Time = 12 months = 1 year.
Do this: Convert at the end and underline the unit.
Usually costs: The final mark; reported for 2024 Q4(i).
Mistake: Keeping the negative root
Time cannot be negative.
Wrong working
$n = 16$ or $n = - 17$ .
Right working
$n = 16$ ; $n = - 17$ is rejected.
Do this: Write 'rejecting the negative value' explicitly.
Usually costs: Half a mark for incomplete reasoning.

Exam tips

In 2024 Q4(i), ₹2,000 a month at 8% earned ₹1,040; the expected working led to $n^{2} + n - 156 = 0$ , $n = 12$ months = 1 year.
Show the quadratic equation in standard form; it earns a method mark even if you factorise wrongly.
For instalment questions, the growth factor (for example 26 or 12.585) can be shown as one line.
Check your $n$ or $P$ by putting it back; it takes 30 seconds.

Practice questions on Monthly Instalment and Time of a Recurring Deposit

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In a recurring deposit, n satisfies $n^{2} + n - 156 = 0$ . The time is:
(a) 12 months
(b) 13 months
(c) 156 months
(d) 6 months
Hint
$12 \times 13 = 156$ .
Answer
(a) 12 months (this equation arose in 2024 Q4(i)).
Q2 · 3 marks · Short answerBoard level
A recurring deposit for 1 year at 6% per annum earns ₹390 interest. Find the monthly instalment.
Hint
$I = P \times 6.5 \times 0.06$ .
Answer
$0.39 P = 390 \Rightarrow P = ₹1, 000$ .
Q3 · 3 marks · Short answerHarder board level
A recurring deposit for 3 years at 7.5% per annum earns ₹8,325 interest. Find the monthly instalment.
Hint
$\frac{36 \times 37}{24} = 55.5$ .
Answer
$P \times 55.5 \times 0.075 = 4.1625 P = 8325 \Rightarrow P = ₹2, 000$ .
Q4 · 3 marks · Short answerHarder board level
Anita deposits ₹2,000 per month at 6% per annum and gets ₹3,000 as interest. Find the time in years.
Hint
$n (n + 1) = \frac{2400 \times 3000}{2000 \times 6}$ .
Answer
$n (n + 1) = 600 \Rightarrow n = 24$ months $= 2$ years.
Q5 · 3 marks · Short answerHarder board level
A recurring deposit for 18 months at 8% per annum matures to ₹19,140. Find the monthly instalment.
Hint
Growth factor $= 18 + 14.25 \times 0.08$ .
Answer
$18 + 1.14 = 19.14$ ; $P = \frac{19140}{19.14} = ₹1, 000$ .
Q6 · 4 marks · Short answerHarder board level
Sunil deposits ₹1,000 a month at 12% per annum and receives ₹12,780 at maturity. Form a quadratic equation in n and find the time.
Hint
$1000 n + 5 n (n + 1) = 12780$ .
Answer
$5 n^{2} + 1005 n - 12780 = 0 \Rightarrow n^{2} + 201 n - 2556 = 0 \Rightarrow (n - 12) (n + 213) = 0$ , so $n = 12$ months (1 year).
Q7 · 4 marks · Short answerHOTS / competency
Two friends each deposit the same monthly amount at 8% per annum. Asha's deposit runs for 1 year and Bina's for 2 years. Bina's interest is ₹1,110 more than Asha's. Find the monthly instalment.
Hint
Difference $= P \times (25 - 6.5) \times 0.08$ .
Answer
$1.48 P = 1110 \Rightarrow P = ₹750$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Solve $n (n + 1) = 420$ .Show answer
$n = 20$ .
2.2 years at 8%: interest equals what multiple of P?Show answer
$25 \times 0.08 = 2$ , so $I = 2 P$ .
3.Interest ₹1,300 for 1 year at 10%. Instalment?Show answer
$0.65 P = 1300 \Rightarrow P = ₹2, 000$ .
4.Write $n (n + 1)$ in terms of I, P and r.Show answer
$n (n + 1) = \frac{2400 I}{P r}$ .
5.Why is one root of the time equation rejected?Show answer
It is negative, and time cannot be negative.

Monthly Instalment and Time of a Recurring Deposit: frequently asked questions

How do I find the time period of a recurring deposit?

Substitute the interest, instalment and rate in the formula to get n(n + 1) = 2400I ÷ (Pr). Solve this quadratic, reject the negative root, and convert months to years if asked.

How do I find the monthly instalment if only the maturity value is given?

Write MV = P × n + P × n(n + 1)/24 × r/100. Both terms contain P, so take P common and divide the maturity value by the bracket.

What if n(n + 1) does not give a whole number?

In a board question it always will. If it does not, recheck the substitution, especially the r/100 and the 24 in the formula.