Arithmetic and Geometric Progression | ICSE Class 10 Maths Notes

This note covers sequences and series, arithmetic and geometric progressions, common differences and ratios, general terms, sums of a specified number of terms, finding missing quantities, and applications involving salaries, production, savings and repeated growth.

What do sequences, terms and series mean?

A sequence is an ordered arrangement of numbers following a rule. Each number in the arrangement is a term. Its position matters: the first term, second term and third term refer to particular places, rather than to the sizes of the numbers.

Write a for the first term and aₙ for the term in position n, where n is a positive integer, meaning a whole number greater than zero. Thus a₁ is the first term, a₂ the second, and a₃ the third.

How is a term different from a sum?

A series is the expression formed by adding the terms of a sequence. The sum of that series is the resulting number. For example, 1, 3, 5, 7 is a sequence; 1 + 3 + 5 + 7 is its associated series, with sum 16.

The notation Sₙ means the sum of the first n terms. Therefore, aₙ asks for one term, whereas Sₙ includes every term from the first through the nth. The subscript, the small character below the usual line, identifies the position or number of terms.

Definition: A progression is a sequence whose terms follow a specific pattern. In an arithmetic progression, abbreviated AP, successive terms differ by a fixed amount. In a geometric progression, abbreviated GP, successive non-zero terms have a fixed ratio.

When does a progression have a last term?

A finite sequence contains a fixed, finite number of terms and has a last term. An infinite sequence continues without a last term. Three dots, written …, indicate continuation. A displayed final term after the dots specifies where a finite list stops.

A general term is a formula describing the term in position n. It lets you calculate a distant term without listing all the preceding ones. Even when the underlying progression continues indefinitely, Sₙ refers here to the sum of a finite first n terms.

How do you recognise an arithmetic progression?

In an arithmetic progression, each term after the first is obtained by adding the same fixed number to its predecessor. This fixed number is the common difference, denoted by d. Calculate it by subtracting the earlier term from the immediately following term.

d = a₂ − a₁. The same difference must hold between every pair of consecutive terms, meaning terms next to one another. Knowing that a list is an AP allows you to use any one consecutive pair to find d.

Can the common difference be negative or zero?

The common difference can be positive, negative or zero. A negative difference means that adding d decreases the value. A zero difference leaves the value unchanged, so a constant sequence is an AP. The first term and the common difference together determine its terms.

Worked example 1. Identify the first term and common difference of the AP 3/2, 1/2, −1/2, −3/2, … .

Answer: a = 3/2 and d = 1/2 − 3/2 = −1. Checking the next difference gives −1/2 − 1/2 = −1. The negative sign belongs to the difference and must remain in later substitutions.

What does a decreasing AP look like?

A ladder with rung lengths 45, 43, 41, 39, 37, 35, 33 and 31 centimetres, measured from bottom to top, gives a finite AP. The abbreviation cm means centimetres. Each rung is 2 cm shorter than the one below it, so d = −2 cm.

The drawing shows two side rails joined by horizontal rungs. The rails are farther apart at the bottom, and the rungs become shorter towards the top.

See Fig. 5.1 in your NCERT textbook

Do not confuse a visible pattern with an AP. In −2, 2, −2, 2, …, the successive differences include 4 and −4. Since these differ, the sequence fails the constant-difference test even though the signs repeat regularly.

How is the general term of an AP obtained?

Starting with first term a and adding common difference d repeatedly gives a, a + d, a + 2d, a + 3d, … . The second term needs one addition of d, the third needs two, and the fourth needs three.

Result: General term of an AP

aₙ = a + (n − 1)d. Here aₙ is the nth term, a the first term, d the common difference, and n the positive integer giving the term's position. The multiplier n − 1 counts the steps after the first term.

To find a particular term, identify a and d, replace n with the required position, and evaluate the expression. Brackets around n − 1 preserve the order of calculation. For the first term, n = 1 makes the added part zero, as required.

Worked example 2. Find the tenth term of the AP 2, 7, 12, … .

Answer: a = 2, d = 7 − 2 = 5 and n = 10. Hence a₁₀ = 2 + (10 − 1) × 5 = 2 + 45 = 47. There are nine equal increases between the first and tenth terms.

How do you find the position of a known term?

If a term's value is given, place that value on the left of the general-term equation and solve for n. The answer must be a positive integer. A fractional value of n cannot identify a term position in the progression.

Worked example 3. Which term of the AP 21, 18, 15, … is −81?

Answer: a = 21 and d = −3. Therefore −81 = 21 + (n − 1)(−3) = 24 − 3n. This gives 3n = 105 and n = 35. Thus −81 is the thirty-fifth term.

Membership is a separate question from numerical size. For the AP 5, 11, 17, 23, …, setting aₙ = 301 gives 301 = 5 + 6(n − 1), hence n = 151/3. This is not a positive integer, so 301 is not a term.

How can given terms determine an AP?

Two terms at known, different positions provide two equations involving a and d. Substitute each position into the general-term formula. Subtracting the equations removes a, leaving an equation for d. Then substitute the common difference back to find the first term.

What changes when the given terms are not consecutive?

The difference between two non-consecutive terms represents several steps of size d. Count the difference between their positions before finding the common difference. For example, the move from the third position to the seventh contains four steps, rather than one.

Worked example 4. Determine the AP whose third term is 5 and seventh term is 9.

Answer: The equations are a + 2d = 5 and a + 6d = 9. Subtracting gives 4d = 4, so d = 1. Substitution into a + 2d = 5 gives a = 3. The required AP is 3, 4, 5, 6, 7, … .

After finding the progression, check both given positions. The third term is 3 + 2 × 1 = 5, and the seventh is 3 + 6 × 1 = 9. This check catches a wrong sign or an incorrect count of steps.

How do you count terms from the end?

For a finite AP, let l denote its last term. Reversing its order gives another AP whose first term is l and whose common difference has the opposite sign. This lets you count from the end using the familiar general-term method.

Worked example 5. Find the eleventh term from the end of the AP 10, 7, 4, …, −62.

Answer: In reverse order the first term is −62 and the common difference is 3. The eleventh term of this reversed AP is −62 + (11 − 1) × 3 = −32. Therefore, the required term is −32.

The original progression contains 25 terms because −62 = 10 + (n − 1)(−3) gives n = 25. Its eleventh term counted backwards is therefore its fifteenth term counted forwards. Counting includes the last term itself as the first position from the end.

How do you find the sum of the first n terms of an AP?

The sum formula adds a chosen number of consecutive terms beginning with the first. It depends on the number of terms, not merely on the number of gaps between them. Keep Sₙ distinct from aₙ throughout the calculation.

Result: Sum of an AP

Sₙ = n[2a + (n − 1)d]/2. Equivalently, if l is the nth term and the last term being included, Sₙ = n(a + l)/2. The second form is useful when the first and last terms and the number of terms are known.

Why does pairing terms produce the sum formula?

1. Write Sₙ as a + (a + d) + … + [a + (n − 1)d].
2. Write the same sum in reverse order, beginning with a + (n − 1)d and ending with a.
3. Add corresponding entries in the two lines. Every pair has value 2a + (n − 1)d, which is also a + l.
4. There are n such pairs, so 2Sₙ = n[2a + (n − 1)d]. Divide by 2 to obtain the sum.

This argument does not require an even number of terms. It adds two complete copies of the series, with n terms in each copy. Dividing by two then recovers the sum of a single copy.

Worked example 6. Find the sum of the first 22 terms of the AP 8, 3, −2, … .

Answer: a = 8, d = −5 and n = 22. Thus S₂₂ = 22[2 × 8 + (22 − 1)(−5)]/2 = 11(16 − 105) = 11(−89) = −979.

A negative total is possible even when the first term is positive. The later negative terms contribute to the sum as negative quantities. Replacing them with their positive sizes would change the series being added.

For the positive integers 1, 2, 3, …, n, the first term is 1 and the last is n. The formula becomes Sₙ = n(n + 1)/2. It gives 500500 for the sum of the first 1000 positive integers.

How are AP formulae used in applications and inverse problems?

In an inverse problem, the sum or a later term is known and an earlier quantity must be found. Choose the formula that contains the given quantities, solve the resulting equation, and interpret the result in the original situation.

Can a sum determine a later term?

Worked example 7. The first 14 terms of an AP have sum 1050, and its first term is 10. Find its twentieth term.

Answer: 1050 = 14[20 + 13d]/2 = 140 + 91d. Thus d = 10. The twentieth term is a₂₀ = 10 + 19 × 10 = 200. First use the sum to find d, then use the term formula.

Sometimes a prescribed sum permits more than one number of terms. For the AP 24, 21, 18, …, setting Sₙ = 78 gives n² − 17n + 52 = 0. Factorising gives (n − 4)(n − 13) = 0, so both n = 4 and n = 13 work.

Here the terms from the fifth through the thirteenth have total zero. Adding them changes the number of terms without changing the sum. Check each positive integer solution rather than discarding one merely because another has already been found.

How does a fixed yearly increase create an AP?

Worked example 8. A television manufacturer produces 600 sets in the third year and 700 in the seventh. Assuming that the production increases uniformly by a fixed number every year, find first-year production, tenth-year production and total production in the first seven years.

Answer: a + 2d = 600 and a + 6d = 700 give d = 25 and a = 550. Hence a₁₀ = 550 + 9 × 25 = 775 sets, while S₇ = 7[2 × 550 + 6 × 25]/2 = 4375 sets.

The assumption of a fixed increase is what makes the AP model applicable. Production in one specified year is a term; total production over the first seven years is a sum. Both answers count sets, but they describe different quantities.

How do you recognise a GP and find its general term?

A geometric progression is a sequence of non-zero terms in which the ratio of each term to its immediately preceding term is constant. This fixed multiplier is the common ratio, denoted by r. Calculate it as the later term divided by the earlier term.

r = a₂/a₁. The same ratio must hold for every consecutive pair. With first term a, the GP is a, ar, ar², ar³, … . Here ar means a multiplied by r, while r² and r³ mean the second and third powers of r.

Result: General term of a GP

aₙ = arⁿ⁻¹. The exponent n − 1 counts the multiplications by r needed after the first term. An exponent indicates a power: for a positive integer exponent, it counts how many copies of the base are multiplied together.

The first term requires no multiplication by r, giving ar⁰ = a because r is non-zero. The second term is ar, the third ar² and the fourth ar³. This parallels the AP's step count, although the operation is multiplication rather than addition.

Worked example 9. Find the tenth and general terms of the GP 5, 25, 125, … .

Answer: a = 5 and r = 25/5 = 5. Therefore a₁₀ = 5 × 5⁹ = 5¹⁰, and aₙ = 5 × 5ⁿ⁻¹ = 5ⁿ. Keep the first factor 5 when combining the powers.

Can ratios be fractional or negative?

A GP need not increase. The sequence 0.01, 0.0001, 0.000001, … has common ratio 0.01. Multiplication by this positive fraction makes each positive term smaller. The multiplier remains fixed even though the amount subtracted between consecutive terms does not.

The GP 1/9, −1/27, 1/81, −1/243, … has common ratio −1/3. Its signs alternate because multiplying by a negative number changes the sign. Retain brackets around a negative ratio when raising it to a power.

Note: The ratio definition used here requires non-zero terms. Dividing by a zero preceding term would be undefined. A ratio of 1 is allowed and gives a constant non-zero progression.

How do you find missing quantities in a GP?

Write each given term using aₙ = arⁿ⁻¹. If the first term and ratio are known, finding a term position becomes an equation involving powers. Expressing both sides using the same base can reveal the required exponent and hence the position.

How is a term position recovered from a power?

Worked example 10. Which term of the GP 2, 8, 32, … is 131072?

Answer: a = 2 and r = 4. Setting 131072 = 2 × 4ⁿ⁻¹ gives 65536 = 4ⁿ⁻¹. Since 65536 = 4⁸, we have n − 1 = 8 and n = 9. The number is the ninth term.

The exponent and the position differ by one. In this example, recognising the eighth power of 4 does not make the answer the eighth term. The extra first term is represented by the factor a before the repeated multiplications begin.

Why is division useful when two terms are known?

Two given terms provide equations containing a common factor a. Dividing one equation by the other removes that factor. The difference between the term positions becomes the exponent of r. Once the ratio is found, substitute into either original equation.

Worked example 11. A GP has third term 24 and sixth term 192. Find its tenth term.

Answer: ar² = 24 and ar⁵ = 192. Dividing gives r³ = 192/24 = 8, so r = 2. Then a = 24/2² = 6. Therefore a₁₀ = 6 × 2⁹ = 3072.

Check the two original conditions before finding the requested term: 6 × 2² = 24 and 6 × 2⁵ = 192. These confirm both the first term and ratio. The quotient 192/24 is 8, but this equals r³ because the positions are three steps apart.

For both types of progression, distinguish a term value from a term position. Values may be fractions or negative numbers; positions count entries and are positive integers. Interchanging these roles produces equations that describe a different question.

How do you calculate the sum of the first n terms of a GP?

For a GP, the sum of its first n terms is Sₙ = a + ar + ar² + … + arⁿ⁻¹. Multiplication of the complete expression by r shifts the powers by one. Subtracting the shifted expression makes all its middle terms cancel.

Result: Finite geometric sum

When r ≠ 1, meaning r is not equal to 1, Sₙ = a(1 − rⁿ)/(1 − r). An equivalent form is Sₙ = a(rⁿ − 1)/(r − 1). Change the signs in both numerator and denominator when moving between them.

How is the GP sum formula derived?

1. Write Sₙ = a + ar + ar² + … + arⁿ⁻¹.
2. Multiply by r to get rSₙ = ar + ar² + … + arⁿ.
3. Subtract the second expression from the first: (1 − r)Sₙ = a − arⁿ = a(1 − rⁿ).
4. For r ≠ 1, divide by 1 − r to obtain Sₙ = a(1 − rⁿ)/(1 − r).

For r = 1, every term equals a, so Sₙ = na. This case needs its own formula because the denominator 1 − r in the other expression would be zero. The general-term formula itself remains valid.

Worked example 12. Find the sum of the first n terms and first five terms of 1 + 2/3 + 4/9 + … .

Answer: a = 1 and r = 2/3. Hence Sₙ = [1 − (2/3)ⁿ]/[1 − 2/3] = 3[1 − (2/3)ⁿ]. Thus S₅ = 3(1 − 32/243) = 3 × 211/243 = 211/81.

Worked example 13. How many terms of the GP 3, 3/2, 3/4, … have sum 3069/512?

Answer: Here a = 3 and r = 1/2, so 3069/512 = 6[1 − (1/2)ⁿ]. Dividing by 6 and rearranging gives (1/2)ⁿ = 1/1024. Since 1024 = 2¹⁰, the required number is n = 10.

Keep fractional powers exact until the calculation is complete. In the first example, rounding 2/3 prematurely would alter the fifth power and the final sum. These formulae add a specified finite number of terms, even when more terms could follow.

How do you choose between AP and GP in a word problem?

Identify the operation connecting successive quantities. A fixed increase or decrease suggests an AP; multiplication by a fixed factor suggests a GP. Then identify whether the question asks for a single quantity at a stated stage or an accumulated sum.

How do the two models compare?

A monthly salary starting at ₹8000 and increasing by ₹500 each year follows an AP across the years. The symbol ₹ denotes rupees. The fifth year's monthly salary is 8000 + 4 × 500 = ₹10000, rather than a sum of five salaries.

How should you handle an initial value at time zero?

Worked example 14. A culture starts with 30 bacteria and its number doubles every hour. Find the number at the end of the second hour, fourth hour and nth hour.

Answer: There are n doublings after n hours, so the number is 30 × 2ⁿ. At the end of the second hour it is 30 × 2² = 120; at the end of the fourth hour it is 30 × 2⁴ = 480.

Here n counts completed hours. The original 30 bacteria occur before the first doubling. If the original count is labelled as the first term of the GP, the count after n hours occupies position n + 1. This explains why the time formula uses exponent n.

For a savings scheme in which ₹8000 becomes 5/4 times its previous amount every three years, the amounts after 3, 6, 9 and 12 years are ₹10000, ₹12500, ₹15625 and ₹19531.25. These amounts form a GP with ratio 5/4.

State exactly where counting begins. In that savings list, the first listed term is the amount after three years; in the bacteria example, the original amount is at time zero. Correct indexing connects the formula to the actual quantity being asked for.

Glossary

• Sequence — An ordered arrangement of numbers generated according to a specified rule.
• Term — An individual number occupying a particular position in a sequence.
• Progression — A sequence whose successive terms follow a specific mathematical pattern.
• Arithmetic progression — A sequence in which each term after the first is obtained by adding a fixed number.
• Common difference — The fixed number added to each preceding term of an arithmetic progression.
• Geometric progression — A sequence of non-zero terms with a constant ratio between each term and its predecessor.
• Common ratio — The fixed multiplier connecting consecutive terms in a geometric progression.
• General term — A formula that gives a sequence's term from its position number.
• Finite sequence — A sequence containing a fixed number of terms and having a last term.
• Infinite sequence — A sequence that continues without ending and therefore has no last term.
• Series — The expression obtained by adding the terms of a sequence in order.
• Sum of a series — The numerical result obtained when the indicated terms of the series are added.

Common errors and misconceptions

• Misconception: The common difference is the earlier term minus the later one. Correct: Subtract the earlier term from the immediately following term, retaining the resulting sign.
• Misconception: An AP must increase. Correct: Its common difference may be negative or zero, so decreasing and constant sequences can also be arithmetic progressions.
• Misconception: The nth AP term is a + nd. Correct: It is a + (n − 1)d, because the first term is already present before any additions.
• Misconception: A fractional answer for n identifies an approximate term position. Correct: Positions must be positive integers; a fractional result means the proposed value is not a term.
• Misconception: The GP sum formula works unchanged when r = 1. Correct: Use Sₙ = na, since division by 1 − r would otherwise divide by zero.
• Misconception: Dividing two given GP terms directly gives r. Correct: It gives r only for consecutive terms; terms several positions apart produce a corresponding power of r.
• Misconception: A term and a sum answer the same question. Correct: aₙ gives one term, while Sₙ adds all terms from the first through the nth.

Exam-style questions with model answers

Q1. Find the first term and common difference of the AP 3/2, 1/2, −1/2, −3/2, … . [2 marks]

1. The first term is a = 3/2, because this is the first number in the given progression.
2. The common difference is d = 1/2 − 3/2 = −1, found by subtracting the first term from the second.

Q2. Find the tenth term of the AP 2, 7, 12, …, showing the formula and substitution. [3 marks]

1. The first term is a = 2 and the common difference is d = 7 − 2 = 5. The required position is n = 10.
2. Use the general-term formula aₙ = a + (n − 1)d, which counts the equal increases after the first term.
3. Substituting gives a₁₀ = 2 + (10 − 1) × 5 = 2 + 45 = 47. Therefore, the tenth term is 47.

Q3. Determine the first term and common difference of an AP whose third term is 5 and seventh term is 9. Write its first five terms. [4 marks]

1. For the third term, the general-term formula gives a + 2d = 5, where a is the first term and d is the common difference.
2. For the seventh term, the corresponding equation is a + 6d = 9.
3. Subtract the first equation from the second: 4d = 4. Hence d = 1.
4. Substitute into a + 2d = 5 to get a = 3. The first five terms are 3, 4, 5, 6 and 7.

Q4. A manufacturer produces 600 television sets in the third year and 700 in the seventh year. Assuming that production increases uniformly by a fixed number every year, find the annual increase, first-year production, tenth-year production and total production in the first seven years. [5 marks]

1. Let a be first-year production and d the fixed annual increase. The third-year and seventh-year figures give a + 2d = 600 and a + 6d = 700.
2. Subtracting the equations gives 4d = 100, so the annual increase is 25 sets. This is the common difference of the production AP.
3. Using a + 2d = 600 gives a = 600 − 50 = 550 sets in the first year.
4. The tenth-year production is a₁₀ = a + 9d = 550 + 9 × 25 = 775 sets.
5. The total for the first seven years is S₇ = 7[2a + 6d]/2 = 7[1100 + 150]/2 = 4375 sets.

Q5. Find the sum of the first five terms of the GP 1, 2/3, 4/9, …, showing the formula and exact calculation. [3 marks]

1. Here a = 1, r = 2/3 and n = 5. The common ratio is found by dividing the second term by the first.
2. Since r is not 1, use Sₙ = a(1 − rⁿ)/(1 − r). Substitution gives S₅ = [1 − (2/3)⁵]/[1 − 2/3].
3. Simplifying exactly gives S₅ = 3(1 − 32/243) = 3 × 211/243 = 211/81. This adds the first five terms, rather than finding the fifth term alone.

Q6. A GP has third term 24 and sixth term 192. Find its common ratio, first term and tenth term. [4 marks]

1. Let a be the first term and r the common ratio. The given terms yield ar² = 24 and ar⁵ = 192.
2. Dividing the second equation by the first gives r³ = 192/24 = 8, so r = 2.
3. Substitute this ratio into ar² = 24. Thus 4a = 24 and the first term is a = 6.
4. Use a₁₀ = ar⁹ to obtain a₁₀ = 6 × 2⁹ = 3072. The exponent is nine because the first term precedes the nine multiplications.

Q7. A culture initially contains 30 bacteria and the number doubles every hour. Find the number present at the end of the second and fourth hours. [2 marks]

1. After two completed hours there have been two doublings, so the number is 30 × 2² = 120 bacteria.
2. After four completed hours there have been four doublings, giving 30 × 2⁴ = 480 bacteria.

Q8. How many terms of the AP 24, 21, 18, … must be added to obtain 78? Show why each answer is admissible. [5 marks]

1. The first term is a = 24 and the common difference is d = −3. Let n be the positive integer number of terms required.
2. Use the AP sum formula to write 78 = n[2 × 24 + (n − 1)(−3)]/2 = n(51 − 3n)/2.
3. Multiplying by 2 and rearranging gives 3n² − 51n + 156 = 0. Dividing by 3 yields n² − 17n + 52 = 0.
4. Factorising gives (n − 4)(n − 13) = 0, so n = 4 or n = 13. Both are positive integers and therefore possible term counts.
5. Substitution gives S₄ = 4(51 − 12)/2 = 78 and S₁₃ = 13(51 − 39)/2 = 78. Thus both four terms and thirteen terms give the required sum.

Key takeaways

• An AP adds a fixed common difference; a GP multiplies each non-zero term by a fixed common ratio.
• The formula aₙ = a + (n − 1)d finds an AP term by counting additions after the first term.
• The AP sum is Sₙ = n[2a + (n − 1)d]/2, or n(a + l)/2 when the last included term is known.
• The GP general term aₙ = arⁿ⁻¹ contains n − 1 multiplications after the first term, so its exponent differs from its position.
• For a GP with r ≠ 1, use Sₙ = a(1 − rⁿ)/(1 − r); for r = 1, use Sₙ = na.
• Subtract equations for known AP terms to find the difference; divide equations for known GP terms to find a power of the ratio.
• Term positions are positive integers, so check that a solution for n can represent an actual position or number of terms.
• In applications, define the starting stage and decide whether the question requires one term or the sum of several terms.

Test yourself

Why is 2, 2, 2, 2, … an AP?

Each term after the first is obtained by adding zero. The common difference is therefore d = 0, which is allowed.

Why is −2, 2, −2, 2, … not an AP?

The first two consecutive differences are 4 and −4. Since the difference is not constant, the sequence is not an AP.

Why does the AP general-term formula contain n − 1?

The first term is already present. Reaching position n therefore requires n − 1 additions of the common difference after it.

Can 301 be a term of the AP 5, 11, 17, 23, …?

No. The equation 301 = 5 + 6(n − 1) gives n = 151/3, which is not a positive integer position.

What is the common ratio of 1/9, −1/27, 1/81, −1/243, …?

The common ratio is −1/3. Dividing each term by its predecessor gives this same negative multiplier, which makes the signs alternate.

Why does a GP with r = 1 need a separate sum formula?

The denominator 1 − r becomes zero. Instead, add n equal terms directly to obtain the valid formula Sₙ = na.

What is the difference between aₙ and Sₙ?

The symbol aₙ denotes the single term in position n, whereas Sₙ denotes the sum of all terms from the first through that position.

Why is the count after n hourly doublings of an initial 30 bacteria equal to 30 × 2ⁿ?

There are n completed doublings after the initial count. The initial 30 occurs before these multiplications, so the exponent counting them is n.