Quantitative Aptitude · Sequence and Series

Applications of AP and GP: Growth, Depreciation and Compound Interest

Q: Why is compound interest a GP?

After each year the amount is multiplied by the same factor (1 + i). The amounts P, P(1 + i), P(1 + i)², … therefore have a constant ratio, which is the definition of a GP.

Q: Which depreciation method is an AP and which is a GP?

Straight-line depreciation subtracts the same amount every year, so the values form an AP. Reducing balance depreciation takes a fixed percentage of the current value, so the values form a GP with ratio (1 − d).

Q: How do I decide whether to use Tₙ or Sₙ?

Use Tₙ when the question asks for the value at one point in time, such as the population after 5 years. Use Sₙ when it asks for a total, such as the total of all deposits with interest.

Q: How do I handle half-yearly compounding with GP?

Use the rate per half-year, which is the annual rate ÷ 2, and the number of half-years, which is years × 2. The common ratio is then 1 + (annual rate ÷ 2).

Updated 1 October 2026 · Fact-checked

Growth, depreciation and compound interest problems are GP problems: each period the value is multiplied by a fixed factor. Simple interest and straight-line depreciation are AP problems: a fixed amount is added or subtracted each period. Identify the pattern, find a, r or d and n, then apply the nth term or sum formula.

Understand Applications: Growth, Depreciation and Compound Interest

A sequence is a list of numbers that follows a rule. In business, the rule is usually 'add the same amount each period' or 'multiply by the same factor each period'. The first is an AP (arithmetic progression). The second is a GP (geometric progression).

Simple interest is an AP. Interest is the same every year, so the amount grows by a fixed sum. Straight-line depreciation is also an AP, because the same amount is cut from the asset value every year. A savings plan where you raise your deposit by a fixed amount each month is an AP too.

Compound interest is a GP. Each year the amount is multiplied by (1 + i), where i is the rate per period as a decimal. Population growth at a fixed percentage works the same way. Reducing balance depreciation is a GP with a common ratio below 1, namely (1 − d).

The key test is simple. If the question says 'by ₹X each year', think AP. If it says 'by X% each year', think GP. Percentages always compound on the changed value, so they give a GP.

A regular deposit plan with compound interest needs the sum of a GP. Each deposit earns interest for a different number of periods, so the deposit values form a GP. Reading carefully whether deposits are made at the start or the end of each year decides the first term.

Key formulas to remember

nth term of an AP: Tₙ = a + (n − 1)d; a is the first term, d is the common difference, n is the number of terms.
Sum of n terms of an AP: Sₙ = (n ÷ 2) × [2a + (n − 1)d]; Use it for savings plans where the deposit rises by a fixed amount.
nth term of a GP: Tₙ = a × r^(n − 1); r is the common ratio. The first term is a, not a × r.
Sum of n terms of a GP: Sₙ = a(rⁿ − 1) ÷ (r − 1) for r > 1; Sₙ = a(1 − rⁿ) ÷ (1 − r) for r < 1; Both forms are equal. Pick the one with a positive denominator. Not valid for r = 1.
Simple interest: I = P × i × n; A = P(1 + n × i); Yearly amounts form an AP with common difference P × i.
Compound interest: A = P(1 + i)ⁿ; CI = A − P; i is the rate per period as a decimal. Yearly amounts form a GP with ratio (1 + i).
Growth at a fixed percentage: Pₙ = P₀(1 + g)ⁿ; Use for population, sales or production growing g per period.
Reducing balance depreciation: Vₙ = C(1 − d)ⁿ; C is the cost, d is the yearly rate as a decimal. Values form a GP with ratio (1 − d).
Straight-line depreciation: Vₙ = C − n × D, where D = (C − scrap value) ÷ life; Values form an AP with common difference −D.
Sum of an infinite GP: S∞ = a ÷ (1 − r), valid only when |r| < 1; Used when a repeating, shrinking amount continues indefinitely.

How to solve Applications: Growth, Depreciation and Compound Interest questions

Use this method for any word problem that asks you to apply AP or GP to a business situation.

1Read the rule of change. 'By ₹X' means AP. 'By X%' means GP.
2Write the first term a and the common difference d or ratio r. For a percentage increase, r = 1 + rate. For a decrease, r = 1 − rate.
3Find n carefully. Check whether the question counts terms or periods, and whether the starting value is the 0th or the 1st term.
4Decide what is asked: one value (use Tₙ) or a total (use Sₙ).
5For compound interest on a single sum, use A = P(1 + i)ⁿ. Convert the rate to the right period first, for example half-yearly rate = annual rate ÷ 2 and n = years × 2.
6For regular deposits, list each deposit's growth period. Start-of-year deposits earn n, n−1, …, 1 years. End-of-year deposits earn n−1, …, 0 years. Then sum the GP.
7Substitute in the formula and compute. Keep the factor (1 + i)ⁿ exact until the last step.
8Check that your answer is sensible: growth must exceed the start, depreciation must fall below the cost.

Quickest way: Multiplier and option-check method

When to use it: Use it in MCQs on compound interest, growth and depreciation, where the options are far apart or differ in their last digits.

Convert the problem to one multiplier: 1.10 for +10%, 0.90 for −10%, 1.05 for +5%.
Raise it to the power n using small steps. Remember 1.1² = 1.21, 1.1³ = 1.331, 0.9² = 0.81, 0.9³ = 0.729, 1.05² = 1.1025.
Eliminate options first. Growth answers must be above the principal, and depreciation answers below the cost.
For a single-sum question, check the last digit of the product against the options.
For a deposit plan, compute the end-of-year total first. Start-of-year total = end-of-year total × (1 + i). If two options differ by that factor, you know which case applies.
Skip any question that needs a power of 1.0x beyond 4 or 5 with no given table. Each wrong answer costs 0.25 marks, so skip if you are only guessing.

Common mistakes in Applications: Growth, Depreciation and Compound Interest

Using AP for percentage growth, such as adding 10% of the original value each year.
Students confuse simple interest with compound interest.
Fix: Percentage change on the new value each period is a GP. Only 'fixed rupee amount' or 'simple interest' gives an AP.
Using r = 10 or r = 0.10 instead of r = 1.10 for a 10% increase.
The rate is taken as the ratio directly.
Fix: Common ratio = 1 + rate for growth and 1 − rate for decay. Write it as a decimal multiplier before you do anything else.
Using n incorrectly, such as n = 3 for the amount after 2 years because three values are listed.
Students count terms and periods as the same thing.
Fix: After n years the amount is P(1 + i)ⁿ. If you list the starting value as the first term T₁, then the value after n years is T₍ₙ₊₁₎. For example, the value after 2 years is T₃.
Not changing the rate and the number of periods for half-yearly or quarterly compounding.
The annual rate is used by habit.
Fix: Divide the annual rate by the number of compounding periods per year and multiply n by the same number.
Treating start-of-year and end-of-year deposits the same.
The first-term growth period is not checked.
Fix: For end-of-year deposits the first deposit earns n−1 years of interest. For start-of-year deposits it earns n years. Write out the GP before you use the formula.
Applying the sum formula with r < 1 using (rⁿ − 1) ÷ (r − 1) and a sign slip.
Students memorise one form and mishandle negatives.
Fix: Use (1 − rⁿ) ÷ (1 − r) when r < 1 so that the numerator and denominator are both positive.

Worked examples

Example 1

A machine costing ₹1,00,000 depreciates at 10% per year on the reducing balance method. Its value at the end of 3 years is: (A) ₹70,000 (B) ₹72,900 (C) ₹73,000 (D) ₹90,000

Show the solution

The rate is a percentage, so the values form a GP.
Cost C = ₹1,00,000, d = 0.10, so r = 1 − 0.10 = 0.9.
Value after 3 years = 1,00,000 × 0.9³.
0.9² = 0.81 and 0.9³ = 0.729.
Value = 1,00,000 × 0.729 = ₹72,900.
Check: ₹70,000 is the value under straight-line depreciation at 10% of original cost per year (₹30,000 in total), ₹90,000 is the value after only 1 year, and ₹73,000 is a close distractor that does not match the calculation. Only ₹72,900 fits.

Answer: (B) ₹72,900

Example 2

Meera saves ₹1,000 in the first month. Each following month she saves ₹200 more than the previous month. Her total savings in 12 months are: (A) ₹24,000 (B) ₹25,200 (C) ₹26,400 (D) ₹28,800

Show the solution

The deposit rises by a fixed amount, so this is an AP.
a = 1,000, d = 200, n = 12.
Sₙ = (n ÷ 2) × [2a + (n − 1)d].
S₁₂ = 6 × [2,000 + 11 × 200].
11 × 200 = 2,200, so the bracket = 2,000 + 2,200 = 4,200.
S₁₂ = 6 × 4,200 = ₹25,200.

Answer: (B) ₹25,200

Example 3

Rohan deposits ₹1,000 at the start of each year for 3 years. Interest is 10% per year compounded annually. The total value at the end of year 3 is: (A) ₹3,000 (B) ₹3,310 (C) ₹3,641 (D) ₹3,300

Show the solution

Each deposit grows by compound interest, so use a GP.
The first deposit earns 3 years of interest: 1,000 × 1.1³ = 1,331.
The second deposit earns 2 years: 1,000 × 1.1² = 1,210.
The third deposit earns 1 year: 1,000 × 1.1 = 1,100.
Total = 1,331 + 1,210 + 1,100 = ₹3,641.
Check with the formula: a = 1,100, r = 1.1, n = 3. S = 1,100 × (1.1³ − 1) ÷ 0.1 = 1,100 × 0.331 ÷ 0.1 = 1,100 × 3.31 = ₹3,641.
The value ₹3,310 is the end-of-year deposit total, which is the usual trap.

Answer: (C) ₹3,641

Exam tips

Spot the pattern first. 'Fixed rupees' means AP and 'percentage' means GP. This decides the whole question in seconds.
Memorise 1.1³ = 1.331, 0.9³ = 0.729, 1.05² = 1.1025 and 1.2² = 1.44. They appear again and again in options.
Watch for traps in the options. The simple interest answer and the end-of-year deposit answer are often placed as wrong choices.
Check the compounding period (half-yearly, quarterly) before you calculate. It is the most common hidden twist.
If a question needs a high power with no log or table given, skip it. A wrong answer costs 0.25 marks.

Practice questions from Sequence and Series

Applications: Growth, Depreciation and Compound Interest: frequently asked questions

Why is compound interest a GP?

After each year the amount is multiplied by the same factor (1 + i). The amounts P, P(1 + i), P(1 + i)², … therefore have a constant ratio, which is the definition of a GP.

Which depreciation method is an AP and which is a GP?

Straight-line depreciation subtracts the same amount every year, so the values form an AP. Reducing balance depreciation takes a fixed percentage of the current value, so the values form a GP with ratio (1 − d).

How do I decide whether to use Tₙ or Sₙ?

Use Tₙ when the question asks for the value at one point in time, such as the population after 5 years. Use Sₙ when it asks for a total, such as the total of all deposits with interest.

How do I handle half-yearly compounding with GP?

Use the rate per half-year, which is the annual rate ÷ 2, and the number of half-years, which is years × 2. The common ratio is then 1 + (annual rate ÷ 2).