Sequence and Series | ISC Class 11 Maths Notes

This note covers sequences and series, arithmetic and geometric progressions, general terms, finite sums, infinite geometric sums, arithmetic and geometric means, the relationship between the means, special sums of natural numbers, and the method of differences.

What is a sequence, and how are its terms described?

Definition: A sequence is an arrangement of numbers in a definite order according to a rule. Each number in that arrangement is a term.

Write the terms as a₁, a₂, a₃, …, aₙ, …. Here n is a natural number, meaning a positive integer, and the subscript gives a term's position. The symbol aₙ means the nth term, also called the general term. The dots indicate continuation according to the rule.

A finite sequence contains a fixed, finite number of terms. An infinite sequence continues without a final term. For example, the even natural numbers form the infinite sequence 2, 4, 6, …, with aₙ = 2n.

How does a formula generate terms?

Often, a rule can be expressed by an algebraic formula. Substitute the required position into that formula. The position and the value are different: n identifies which term is wanted, while aₙ gives the number at that position.

Worked example 1. Write the first three terms of the sequence aₙ = 2n + 5.

Answer: Substitute n = 1, 2 and 3 in order. Then a₁ = 2(1) + 5 = 7, a₂ = 2(2) + 5 = 9 and a₃ = 2(3) + 5 = 11. The required sequence begins 7, 9, 11.

What is a recurrence relation?

A recurrence relation generates terms from earlier terms, together with specified starting values. The Fibonacci sequence starts with a₁ = a₂ = 1 and uses aₙ = aₙ₋₁ + aₙ₋₂ for n > 2. It begins 1, 1, 2, 3, 5, 8, ….

Not every sequence should be expected to have its terms given by a specific formula. A rule for generating successive terms is still needed. A sequence may also be viewed as a function whose inputs are natural-number positions and whose outputs are the corresponding terms.

How does a series differ from a sequence?

A series is the indicated addition of the terms of a sequence. Thus, a₁, a₂, …, aₙ is a sequence, while a₁ + a₂ + … + aₙ is its associated finite series. Commas list terms; plus signs indicate their addition.

The sum of a series is the number obtained by carrying out that addition. For example, 1 + 3 + 5 + 7 is a series with four terms, and its sum is 16. The expression and its evaluated value have different roles.

How is sigma notation read?

The symbol ∑, the Greek capital letter sigma, means summation. In $\sum_{k=1}^{n} a_k$, k is the summation index, or counter, taking the integer values from 1 to n. The lower and upper limits specify where the addition starts and stops.

Write Sₙ for the sum of the first n terms. Then Sₙ = a₁ + a₂ + … + aₙ. For an infinite sequence, Sₙ is called a partial sum because it includes only the initial n terms.

Worked example 2. A sequence has a₁ = 1 and aₙ = aₙ₋₁ + 2 for n ≥ 2. Find its first five terms and write their associated finite series.

Answer: Starting from 1, add 2 at each step to obtain 1, 3, 5, 7, 9. The series formed from these five terms is 1 + 3 + 5 + 7 + 9.

A finite sum has a specified number of terms. An infinite series has no final term, so assigning it a finite sum requires an additional condition. For geometric series, this condition will be expressed using the common ratio.

How do you recognise an arithmetic progression and find its general term?

An arithmetic progression, abbreviated A.P., is a sequence in which every term after the first is obtained by adding the same fixed number to the preceding term. This fixed number is the common difference, denoted by d. It can be positive, negative or zero.

Let a denote the first term. The general form is a, a + d, a + 2d, a + 3d, …. To calculate d, subtract a term from the term immediately after it. Reversing that subtraction changes its sign.

Result: The general term of an A.P.

Use Tₙ as another notation for the nth term. There are n − 1 equal steps from the first term to the nth term, so Tₙ = a + (n − 1)d. The multiplier of d counts steps, not terms.

Worked example 3. Find the tenth term of the A.P. 2, 7, 12, ….

Answer: The first term is a = 2 and the common difference is d = 7 − 2 = 5. With n = 10, T₁₀ = 2 + (10 − 1)5 = 2 + 45 = 47.

When a term value is given, solve the same formula for its position. The answer must be a positive integer to represent a position in the sequence. A negative or fractional solution does not identify a term.

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

Answer: The conditions give a + 2d = 5 and a + 6d = 9. Subtracting gives 4d = 4, so d = 1. Substitution gives a = 3. The progression is 3, 4, 5, 6, 7, ….

For an already specified A.P., one adjacent pair determines d. When testing whether a list follows an arithmetic pattern, the same difference must hold throughout the stated rule, rather than merely for one pair.

How is the sum of an arithmetic progression obtained?

The sum formula adds the first n terms without listing and adding each term separately. If l denotes the last of these n terms, then l = a + (n − 1)d. This is the last included term, even when the whole progression is infinite.

Result: The finite arithmetic sum

Sₙ = n[2a + (n − 1)d]/2 = n(a + l)/2. Use the first form when a, d and n are known. Use the second when the first term, last included term and number of terms are known.

1. Write Sₙ = a + (a + d) + … + l.
2. Write the same sum in reverse order: Sₙ = l + (l − d) + … + a.
3. Add corresponding entries in the two rows. Each pair has the value a + l.
4. There are n pairs, so 2Sₙ = n(a + l). Divide by 2 and substitute l = a + (n − 1)d.

Worked example 5. Find the sum of the first 22 terms of the A.P. 8, 3, −2, ….

Answer: Here a = 8, d = −5 and n = 22. Therefore S₂₂ = 22[2(8) + 21(−5)]/2 = 11(16 − 105) = −979.

The negative result is consistent with a progression whose later terms are negative. A positive first term does not force a positive sum. Keep the sign of the common difference when substituting, especially inside brackets.

Also distinguish a term from a sum. T₂₂ would give only the twenty-second term. S₂₂ includes that term and all twenty-one terms before it. The question's wording determines which formula is appropriate.

How are arithmetic means inserted between two numbers?

The arithmetic mean, abbreviated A.M., of two numbers is half their sum. If a, b and c are consecutive terms of an A.P., then b − a = c − b. Rearrangement gives 2b = a + c, or b = (a + c)/2.

How do several inserted means work?

Let m be the number of arithmetic means to insert between endpoints a and b. Name the inserted terms A₁, A₂, …, Aₘ, with each subscript recording its position among the inserted means. The full progression is a, A₁, A₂, …, Aₘ, b.

There are m + 2 terms in all and m + 1 intervals between them. Consequently b = a + (m + 1)d, giving d = (b − a)/(m + 1). The kth inserted mean is Aₖ = a + kd, where k runs from 1 to m.

This method separates the number inserted from the total number of terms. Dividing the endpoint difference by m would count the intervals incorrectly. Once d is found, repeatedly add it, including the final step that must reach b.

Which forms simplify unknown arithmetic terms?

For three unknown terms, a − d, a, a + d is a useful symmetric form: the terms are arranged at equal distances around a. Here a is the middle term, rather than the first. Their sum is 3a.

For four terms, use a − 3d, a − d, a + d, a + 3d. Their sum is 4a, and consecutive terms differ by 2d. In this representation d is a parameter controlling the spacing; the actual common difference is 2d.

State what a symbol means before substituting into a formula. A letter used for the centre of a symmetric arrangement must not accidentally be treated as its first term.

How do geometric progressions differ from arithmetic progressions?

A geometric progression, abbreviated G.P., is a sequence of non-zero terms in which the ratio of every term after the first to the preceding term is constant. This constant is the common ratio, denoted by r.

With first term a, the progression is a, ar, ar², ar³, …. The ratio is found by dividing the later term by the preceding one. For example, 2, 4, 8, 16, … has r = 2, while 1/9, −1/27, 1/81, −1/243, … has r = −1/3.

Result: The general term of a G.P.

To reach the nth term from the first, multiply by r a total of n − 1 times. Hence Tₙ = arⁿ⁻¹. The first term corresponds to exponent zero, since r⁰ = 1 for the non-zero ratios used here.

Worked example 6. Find the tenth and general terms of the G.P. 5, 25, 125, ….

Answer: Here a = 5 and r = 25/5 = 5. Thus T₁₀ = 5 × 5⁹ = 5¹⁰ and Tₙ = 5 × 5ⁿ⁻¹ = 5ⁿ.

A negative ratio alternates the signs of consecutive terms. Retain the sign when raising r to a power. A ratio smaller than zero is compatible with a G.P.; the defining requirement is a constant ratio between successive non-zero terms.

How can unknown geometric terms and their positions be found?

When two terms of a G.P. are given, write each using Tₙ = arⁿ⁻¹. Dividing the equation for the later term by the earlier one eliminates a. The difference between the term positions becomes the exponent of r.

Worked example 7. The third term of a G.P. is 24 and the sixth term is 192. Find its tenth term.

Answer: The data give ar² = 24 and ar⁵ = 192. Division yields r³ = 192/24 = 8, so r = 2. Then a = 24/2² = 6, and T₁₀ = 6 × 2⁹ = 3072.

How do you locate a given term?

Set the general term equal to the specified value and solve for n. Rewriting both sides as powers of the same positive base can make the comparison direct. Remember that the exponent in the general-term formula is n − 1.

Worked example 8. Which term of the G.P. 2, 8, 32, … is 131072?

Answer: Here a = 2 and r = 4. The equation 2 × 4ⁿ⁻¹ = 131072 gives 4ⁿ⁻¹ = 65536 = 4⁸. Thus n − 1 = 8 and n = 9. The specified number is the ninth term.

Which forms simplify unknown geometric terms?

Three terms can be written as a/r, a, ar, with a non-zero centre a and non-zero ratio r. Their product is a³. The reverse order ar, a, a/r also forms a G.P., but its successive ratio is 1/r.

The four-term form ar³, ar, ar⁻¹, ar⁻³ has successive ratio r⁻² and product a⁴. Here r⁻¹ means 1/r, and similarly r⁻² = 1/r² and r⁻³ = 1/r³. These forms simplify products while preserving equal successive ratios.

How is the finite sum of a geometric progression derived?

To add a G.P., multiply the sum by its common ratio and subtract. The shifted terms cancel in pairs. This is different from reversing and adding an A.P., because geometric terms change by multiplication.

Result: The finite geometric sum

For r ≠ 1, Sₙ = a(1 − rⁿ)/(1 − r) = a(rⁿ − 1)/(r − 1). The two forms are equal because multiplying both numerator and denominator by −1 does not change the fraction.

1. Write Sₙ = a + ar + ar² + … + arⁿ⁻¹.
2. Multiply by r to obtain rSₙ = ar + ar² + … + arⁿ.
3. Subtract the second equation from the first: (1 − r)Sₙ = a − arⁿ.
4. Factor the right-hand side and divide by 1 − r, which is non-zero when r ≠ 1.

If r = 1, every term equals a, so Sₙ = na. The fractional formula cannot be used directly because its denominator would be zero. Treat this case before substituting values.

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

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

Both fractional forms work for any permitted r ≠ 1. Choosing the form with convenient signs can simplify arithmetic. Neither form by itself turns a finite sum into an infinite sum; that requires a separate condition.

When does an infinite geometric series have a finite sum?

An infinite geometric series has infinitely many terms, but its partial sums may approach a fixed number. To say that they approach that number means they can be made arbitrarily close to it by taking sufficiently many terms.

Result: The infinite geometric sum

The notation |r| means the absolute value, or distance of r from zero. If |r| < 1, equivalently −1 < r < 1, the powers rⁿ approach zero as n increases. Hence Sₙ = a(1 − rⁿ)/(1 − r) approaches a/(1 − r).

Write S∞ for this limiting sum, where ∞ denotes infinity. Thus S∞ = a/(1 − r), provided |r| < 1. A series whose partial sums approach a finite number is called convergent.

This condition concerns the size of the ratio as well as its sign. A negative ratio can satisfy it. Requiring only r < 1 would wrongly admit negative ratios whose absolute values are too large.

For the non-zero-term geometric progressions considered here, |r| ≥ 1 does not give a finite limiting sum. In particular, r = 1 repeats the same non-zero term; r = −1 makes the partial sums alternate instead of approaching one value.

Note: Check |r| < 1 before using the infinite-sum formula. Finding an algebraic value for a/(1 − r) is not itself evidence that an infinite series has that sum.

The distinction is between the number of terms being added and the behaviour of the resulting partial sums. The finite formula retains rⁿ. The infinite formula follows only after showing that this power approaches zero under the stated condition.

How are geometric means inserted and compared with arithmetic means?

The geometric mean, abbreviated G.M., of two positive numbers a and b is G = √(ab). Here G denotes the mean, and √ denotes the non-negative square root. The three terms a, G, b form a G.P.

How are several geometric means inserted?

To insert m positive geometric means G₁, G₂, …, Gₘ between positive endpoints a and b, form a, G₁, G₂, …, Gₘ, b. There are m + 1 multiplication steps, so b = arᵐ⁺¹. Choose the positive root r = (b/a)¹⁄⁽ᵐ⁺¹⁾ and set Gₖ = arᵏ.

Worked example 10. Insert three real numbers between 1 and 256 so that the resulting sequence is a G.P. Include both possible real ratios.

Answer: There are five terms, so 256 = 1 × r⁴. The real ratios are r = 4 and r = −4. For r = 4, insert 4, 16, 64. For r = −4, insert −4, 16, −64. Both arrangements end at 256.

The positive values are the inserted positive geometric means. The negative-ratio solution remains valid when the problem asks for real numbers forming a G.P. Distinguish that wording from the positive square-root definition of a geometric mean.

Result: The relationship between A.M. and G.M.

For positive a and b, write A = (a + b)/2 for their arithmetic mean and G = √(ab) for their geometric mean. Then A − G = (√a − √b)²/2 ≥ 0, because the square of a real number is non-negative.

Consequently A ≥ G, with equality precisely when a = b. To recover two numbers from their means, use a + b = 2A and ab = G² together. These equations encode the given sum and product without losing the positivity condition.

How do special sums simplify related series?

The special sums add consecutive natural numbers, their squares, or their cubes. In the following table, k is the summation index and n is a positive integer. Each sum starts at k = 1 and ends at k = n.

How are the formulae combined?

Expand a term before summing when that exposes powers of the index. Addition can be separated into sums, and a fixed numerical factor can be taken outside a sum. Keep the same starting and ending indices in each resulting expression.

Worked example 11. For the sequence with general term aₙ = n(n + 2), find the sum of its first n terms.

Answer: Its kth term is k(k + 2) = k² + 2k. Hence Sₙ = ∑k² + 2∑k, with both sums running from k = 1 to n. Substitution gives Sₙ = n(n + 1)(2n + 1)/6 + n(n + 1) = n(n + 1)(2n + 7)/6.

The sum of squares is different from the square of a sum. The special identity for cubes says that the sum of the first n cubes equals the square of the sum of the first n natural numbers. It does not state the corresponding claim for squares.

If a series omits initial terms, first calculate the appropriate sum from 1 and subtract the omitted part. The bounds belong to the formula just as much as the algebraic expression does.

How does the method of differences recover a general term?

The method of differences compares consecutive terms to reveal a simpler pattern. A first difference is the later term minus the earlier term. A second difference is the difference between consecutive first differences. A useful pattern must hold throughout the sequence being considered.

How are differences added back?

Let ΔTₖ = Tₖ₊₁ − Tₖ denote the kth first difference; Δ is the difference symbol. Adding these differences from k = 1 to n − 1 cancels the intermediate terms, leaving Tₙ − T₁.

Therefore, for n ≥ 2, Tₙ = T₁ + ∑ΔTₖ, where the sum runs from k = 1 to n − 1. This cancellation is called telescoping. It explains why the initial term and the complete difference rule together determine subsequent terms.

For the sequence aₙ = n(n + 2), its kth first difference is (k + 1)(k + 3) − k(k + 2) = 2k + 3. Since a₁ = 3, adding these differences gives aₙ = 3 + n(n − 1) + 3(n − 1) = n(n + 2).

How can partial sums give individual terms?

There is a related subtraction rule: Tₙ = Sₙ − Sₙ₋₁ for n ≥ 2, while T₁ = S₁. The first n − 1 terms occur in both partial sums and cancel. This identity is useful even without an A.P. or G.P.

Worked example 12. The sum of the first n terms of a sequence is Sₙ = 4n − n². Find the first term and its general term.

Answer: T₁ = S₁ = 4 − 1 = 3. For n ≥ 2, Tₙ = (4n − n²) − [4(n − 1) − (n − 1)²] = 5 − 2n. This formula also gives 3 at n = 1.

Subtract the entire expression for Sₙ₋₁ using brackets. The outer minus sign affects every term inside. Finally check that the resulting formula agrees with the separately calculated first term.

Glossary

• Sequence — An arrangement of numbers in a definite order according to a specified rule.
• Term — An individual number occupying a particular position within a sequence.
• General term — An expression giving the value of a term from its position in a sequence.
• Finite sequence — A sequence containing a fixed, finite number of terms and having a final term.
• Recurrence relation — A rule generating later terms from earlier terms together with specified starting values.
• Series — The indicated addition of the terms belonging to a sequence.
• Partial sum — The sum of the first specified number of terms of a sequence.
• Arithmetic progression — A sequence obtained by repeatedly adding a fixed common difference after its first term.
• Common difference — The fixed value obtained by subtracting an arithmetic term from the immediately following term.
• Arithmetic mean — Half the sum of two numbers, forming the middle term of their three-term arithmetic progression.
• Geometric progression — A sequence of non-zero terms with a constant ratio between each term and its predecessor.
• Common ratio — The fixed quotient obtained by dividing a geometric term by its immediately preceding term.
• Geometric mean — The positive square root of the product of two positive numbers.
• Convergent series — A series whose partial sums approach a finite value as more terms are included.
• Telescoping — Cancellation of intermediate terms when consecutive differences are added together.

Common errors and misconceptions

• Misconception: A sequence and its sum are the same object. Correct: A sequence lists terms, a series indicates their addition, and the sum is the resulting value.
• Misconception: The nth arithmetic term is a + nd. Correct: It is a + (n − 1)d because reaching the nth term requires n − 1 additions after the first.
• Misconception: The common difference must be positive. Correct: It may be positive, negative or zero; subtract the earlier term from the later one.
• Misconception: Inserting m means creates m intervals. Correct: Including both endpoints gives m + 2 terms and m + 1 intervals.
• Misconception: The fractional finite G.P. sum formula can be used when r = 1. Correct: Its denominator vanishes; use Sₙ = na instead.
• Misconception: Every infinite G.P. has sum a/(1 − r). Correct: For the non-zero-term progressions here, a finite infinite sum requires |r| < 1.
• Misconception: The geometric mean of positive numbers includes both square-root signs. Correct: Their G.M. is positive; a problem asking for real inserted terms can separately allow negative ratios.
• Misconception: Squaring the sum of natural numbers gives their sum of squares. Correct: [n(n + 1)/2]² is their sum of cubes; the square sum has a different formula.

Exam-style questions with model answers

Q1. The sequence has general term aₙ = 2n + 5, where n is a positive integer. Find its first three terms. [2 marks]

1. Substitute n = 1 into the given expression: a₁ = 2(1) + 5 = 7.
2. For n = 2 and n = 3, obtain a₂ = 9 and a₃ = 11. The first three terms are 7, 9, 11.

Q2. The third term of an A.P. is 5 and its seventh term is 9. Find the first term and common difference, then write the progression. [4 marks]

1. Let a be the first term and d the common difference. Applying the general-term formula to the third term gives a + 2d = 5.
2. The seventh term similarly gives a + 6d = 9. Subtract the third-term equation from this equation to obtain 4d = 4.
3. Hence d = 1. Substituting in a + 2d = 5 gives a = 3.
4. Starting at 3 and adding 1 repeatedly gives the required progression 3, 4, 5, 6, 7, ….

Q3. Find the sum of the first 22 terms of the A.P. 8, 3, −2, …. [3 marks]

1. The first term is a = 8 and the common difference is d = 3 − 8 = −5. The required number of terms is n = 22.
2. Use Sₙ = n[2a + (n − 1)d]/2. Substitution gives S₂₂ = 22[16 + 21(−5)]/2.
3. Therefore S₂₂ = 11(16 − 105) = 11(−89) = −979. This is the total of all twenty-two included terms, rather than the final term alone.

Q4. A G.P. has third term 24 and sixth term 192. Find its tenth term. [4 marks]

1. Let a denote the first term and r the common ratio. The two given terms yield ar² = 24 and ar⁵ = 192.
2. Divide the sixth-term equation by the third-term equation: r³ = 192/24 = 8. Thus the real common ratio is r = 2.
3. Substitute r = 2 in ar² = 24 to obtain a = 24/4 = 6.
4. The tenth term is T₁₀ = ar⁹ = 6 × 2⁹ = 3072.

Q5. For the geometric series 1 + 2/3 + 4/9 + …, obtain its sum to n terms and hence its sum to five terms. [3 marks]

1. The first term is a = 1 and the common ratio is r = 2/3. Since this ratio differs from 1, the fractional finite-sum formula applies.
2. Substitute in Sₙ = a(1 − rⁿ)/(1 − r) to obtain Sₙ = 3[1 − (2/3)ⁿ].
3. Set n = 5. Then S₅ = 3(1 − 32/243) = 3 × 211/243 = 211/81, the required finite sum.

Q6. Insert three real numbers between 1 and 256 to form a G.P. Find both possible real common ratios and the corresponding inserted numbers. [5 marks]

1. Let r be the common ratio. Including the two endpoints, the progression contains five terms, so its form is 1, r, r², r³, r⁴.
2. The last term must equal 256. Hence r⁴ = 256, which has the two real solutions r = 4 and r = −4.
3. With r = 4, successive multiplication gives the three inserted numbers 4, 16 and 64. The final multiplication gives 256.
4. With r = −4, successive multiplication gives −4, 16 and −64. Multiplying −64 by −4 again gives the required endpoint 256.
5. Thus both sets satisfy the request for real inserted numbers. The positive inserted geometric means are 4, 16 and 64.

Q7. Two positive numbers have arithmetic mean 10 and geometric mean 8. Find the numbers. [5 marks]

1. Let the positive numbers be a and b. Their arithmetic mean gives (a + b)/2 = 10, so their sum is a + b = 20.
2. The geometric mean gives √(ab) = 8. Squaring this equation yields ab = 64, the product of the required numbers.
3. Use the identity (a − b)² = (a + b)² − 4ab. Substitution gives (a − b)² = 400 − 256 = 144.
4. Therefore a − b = 12 or −12. Combining either difference with a + b = 20 gives a = 16, b = 4, or the reverse order.
5. The numbers are 4 and 16. Their arithmetic mean is 20/2 = 10 and their geometric mean is √64 = 8, verifying both given conditions.

Q8. Let a and b be positive real numbers. Define their arithmetic mean A = (a + b)/2 and geometric mean G = √(ab). Prove that A ≥ G and state precisely when equality holds. [5 marks]

1. Begin with the difference of the two defined means: A − G = (a + b)/2 − √(ab).
2. Put both terms over the common denominator 2 to obtain A − G = [a + b − 2√(ab)]/2.
3. Because a and b are positive, their real square roots exist. The numerator factors as (√a − √b)².
4. The square of a real number is non-negative. Dividing it by the positive number 2 preserves that property, so A − G ≥ 0 and therefore A ≥ G.
5. Equality holds exactly when √a − √b = 0, which is equivalent to a = b. Conversely, equal positive numbers clearly give equal arithmetic and geometric means.

Key takeaways

• A sequence lists ordered terms, while its associated series indicates addition; evaluating that addition gives the sum.
• An arithmetic progression adds a constant difference; its nth term requires n − 1 additions after the first term.
• The arithmetic sum is n times the average of the first and last included terms.
• A geometric progression multiplies by a constant ratio; use Tₙ = arⁿ⁻¹ for its general term.
• The finite geometric sum needs a separate case when r = 1; an infinite sum requires |r| < 1.
• Inserting m means between two endpoints creates m + 2 terms and m + 1 intervals.
• For two positive numbers, the arithmetic mean is at least the geometric mean, with equality when the numbers coincide.
• Special sums handle powers of consecutive natural numbers; consecutive partial sums recover individual terms through subtraction.

Test yourself

What is the difference between a series and its sum?

A series indicates the addition of sequence terms; its sum is the value obtained by carrying out that addition.

Why does an A.P. general term use n − 1 rather than n?

From the first term to the nth term there are n − 1 steps, each adding the common difference.

What is the common difference in the A.P. 6, 3, 0, −3, …?

The common difference is −3, found by subtracting the earlier term from the following term: 3 − 6 = −3.

How many intervals result when m arithmetic means are inserted between two endpoints?

There are m + 1 intervals between the m + 2 terms, including the two endpoints.

What finite geometric sum formula applies when the common ratio is 1?

Use Sₙ = na because all n terms equal the first term a.

What condition permits the infinite geometric sum a/(1 − r)?

The ratio must satisfy |r| < 1, meaning it lies strictly between −1 and 1.

When do two positive numbers have equal arithmetic and geometric means?

Equality holds precisely when the two positive numbers are equal, making (√a − √b)² zero.

How is the nth term found from consecutive partial sums?

For n ≥ 2, subtract Sₙ₋₁ from Sₙ. Their shared terms cancel, leaving Tₙ; separately, T₁ = S₁.