Sequence and Series ISC Class 11: AP, GP, Means and Special Sums

What Sequence and Series is about

Weightage: Sequence and Series belongs to the unit Algebra, which CISCE weights at 26 of the 80 marks in the ISC Class XI Mathematics Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Mathematics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Arithmetic progression: nth term and sum

An AP has first term a and common difference d = T₂ − T₁ = T₃ − T₂ = …. The nth term is Tₙ = a + (n − 1)d (n − 1 steps of d after the first term). The sum of n terms, found by writing the series forwards and backwards and adding, is Sₙ = (n/2)[2a + (n − 1)d] = (n/2)(a + l), where l is the last term. Useful facts: Tₙ = Sₙ − Sₙ₋₁; if Sₙ is a quadratic in n with no constant term, the series is an AP; the nth term from the end is l − (n − 1)d.

Worked example

For the AP 7, 11, 15, …, find T₂₀ and S₂₀.

Try it first, then show the solutionSolution

1. a = 7, d = 4: T₂₀ = 7 + 19 × 4 = 83
2. S₂₀ = (20/2)(7 + 83) = 10 × 90 = 900

Answer: T₂₀ = 83, S₂₀ = 900

Arithmetic means and choosing terms in an AP

b is the arithmetic mean (AM) of a and c when a, b, c are in AP: 2b = a + c. To insert n AMs between a and b, form an AP of n + 2 terms with first a and last b: d = (b − a)/(n + 1); the means are a + d, a + 2d, …, a + nd. Their sum is n times the single AM (a + b)/2. In word problems, take three terms as a − d, a, a + d and four as a − 3d, a − d, a + d, a + 3d (common difference 2d): the sum then removes d immediately.

Worked example

Three numbers in AP have sum 27 and product 648. Find them.

Try it first, then show the solutionSolution

1. Take a − d, a, a + d: 3a = 27, a = 9
2. (9 − d)(9)(9 + d) = 648 ⇒ 81 − d² = 72 ⇒ d = ±3

Answer: 6, 9, 12 (or 12, 9, 6)

Geometric progression: nth term and sums

A GP has first term a and common ratio r = T₂/T₁ = T₃/T₂ = … (r ≠ 0). The nth term is Tₙ = arⁿ⁻¹. Multiplying Sₙ by r and subtracting gives Sₙ = a(rⁿ − 1)/(r − 1), convenient for r > 1, or equivalently Sₙ = a(1 − rⁿ)/(1 − r), convenient for |r| < 1; if r = 1, Sₙ = na. When |r| < 1, rⁿ → 0 as n grows, so the sum to infinity is S∞ = a/(1 − r). If |r| ≥ 1, the infinite sum does not exist.

Worked example

For the GP 3, 6, 12, …, find T₈ and S₈. Also find the sum to infinity of 16 + 8 + 4 + ….

Try it first, then show the solutionSolution

1. a = 3, r = 2: T₈ = 3 × 2⁷ = 384
2. S₈ = 3(2⁸ − 1)/(2 − 1) = 3 × 255 = 765
3. 16 + 8 + 4 + …: r = 1/2, S∞ = 16/(1 − 1/2) = 32

Answer: 384; 765; 32

Geometric means and choosing terms in a GP

For positive a and c, b is their geometric mean (GM) when a, b, c are in GP: b² = ac, b = √(ac). To insert n GMs between a and b, form a GP of n + 2 terms: r = (b/a)^{1/(n + 1)}; the means are ar, ar², …, arⁿ, and their product is (√(ab))ⁿ. In word problems take three terms in GP as a/r, a, ar and four as a/r³, a/r, ar, ar³, so that their product is a³ or a⁴.

Worked example

(i) Insert 3 GMs between 2 and 162. (ii) Three numbers in GP have product 216 and sum 21. Find them.

Try it first, then show the solutionSolution

1. (i) 2r⁴ = 162 ⇒ r⁴ = 81 ⇒ r = 3 (taking the positive value): means 6, 18, 54
2. (ii) a/r × a × ar = a³ = 216 ⇒ a = 6
3. 6/r + 6 + 6r = 21 ⇒ 6r² − 15r + 6 = 0 ⇒ 2r² − 5r + 2 = 0 ⇒ r = 2 or 1/2

Answer: (i) 6, 18, 54 (ii) 3, 6, 12 (or 12, 6, 3)

Relation between AM and GM

For positive numbers a and b with A = (a + b)/2 and G = √(ab): A − G = (a + b − 2√(ab))/2 = (√a − √b)²/2 ≥ 0, so A ≥ G, with equality exactly when a = b. Since a + b = 2A and ab = G², a and b are the roots of x² − 2Ax + G² = 0, so a, b = A ± √(A² − G²). AM ≥ GM also gives least values: for x > 0, x + k/x ≥ 2√k.

Worked example

The AM and GM of two positive numbers are 13 and 12. Find the numbers.

Try it first, then show the solutionSolution

1. a + b = 26, ab = 144
2. Numbers = 13 ± √(169 − 144) = 13 ± 5

Answer: 18 and 8

Special sums Σn, Σn², Σn³

Σk = 1 + 2 + … + n = n(n + 1)/2; Σk² = n(n + 1)(2n + 1)/6; Σk³ = [n(n + 1)/2]² (the sum of cubes is the square of the sum of the numbers). The Σk² result can be proved by adding (k + 1)³ − k³ = 3k² + 3k + 1 for k = 1 to n: the left side telescopes to (n + 1)³ − 1. To sum a series whose general term is a polynomial in k, write Tₖ as a combination of k³, k², k and 1, and add the special sums.

Worked example

Find the sum of n terms of 1·2 + 2·3 + 3·4 + …, and evaluate it for n = 10.

Try it first, then show the solutionSolution

1. Tₖ = k(k + 1) = k² + k
2. Sₙ = n(n + 1)(2n + 1)/6 + n(n + 1)/2 = n(n + 1)[(2n + 1) + 3]/6 = n(n + 1)(n + 2)/3
3. n = 10: 10 × 11 × 12/3 = 440

Answer: n(n + 1)(n + 2)/3; 440

nth term by the method of differences

If the differences between consecutive terms form an AP (or a GP), the original sequence's nth term can be found. Write Sₙ = T₁ + T₂ + … + Tₙ and the same sum shifted by one place; subtracting gives Tₙ = T₁ + (sum of the first n − 1 differences). If the first differences are in AP, Tₙ is quadratic in n; if the second differences are constant, check by substituting n = 1, 2, 3. Then sum Tₙ using the special sums.

Worked example

Find the nth term and the sum of n terms of 3 + 7 + 13 + 21 + 31 + ….

Try it first, then show the solutionSolution

1. Differences 4, 6, 8, 10, … form an AP with first term 4 and d = 2
2. Tₙ = 3 + (sum of the first n − 1 differences) = 3 + ((n − 1)/2)[8 + 2(n − 2)] = 3 + (n − 1)(n + 2) = n² + n + 1
3. Sₙ = Σk² + Σk + n = n(n + 1)(2n + 1)/6 + n(n + 1)/2 + n = n(n² + 3n + 5)/3
4. Check: S₂ = 2 × 15/3 = 10 = 3 + 7 ✓

Answer: Tₙ = n² + n + 1; Sₙ = n(n² + 3n + 5)/3

Maths in real life: Sequence and Series

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.

Topics in this chapter

• AM–GM Inequality, Special Sums and the Method of Differences
• Arithmetic and Geometric Progressions and Means

Sequence and Series formula sheet

AP nth term and sum

$T_n=a+(n-1)d;S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)$

Tₙ = Sₙ − Sₙ₋₁.

Inserting n AMs

$d=\frac{b-a}{n+1}$

Single AM: (a + b)/2.

GP nth term and sum

$T_n=a{r}^{n-1};S_n=\frac{a(r^n-1)}{r-1}(r\ne 1)$

Use a(1 − rⁿ)/(1 − r) when |r| < 1.

Infinite GP

$S_{\infty }=\frac{a}{1-r},|r|<1$

Does not exist for |r| ≥ 1.

Inserting n GMs

$r={\left(\frac{b}{a}\right)}^{\frac{1}{n+1}}$

Single GM: √(ab).

AM ≥ GM

$\frac{a+b}{2}\ge \sqrt{ab}(a,b>0)$

Numbers = A ± √(A² − G²).

Special sums

$\sum k=\frac{n(n+1)}{2};\sum k^2=\frac{n(n+1)(2n+1)}{6};\sum k^3={\left[\frac{n(n+1)}{2}\right]}^2$

Σ1 = n.

Sequence and Series: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Sequence and Series

The question patterns to expect, and the method that works for each.

Tricks to remember Sequence and Series

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

• n − 1 steps

  To reach the nth term from the first, you take n − 1 steps: a + (n − 1)d, arⁿ⁻¹.

  Use it when: nth terms.

• Average times count

  Sum of an AP = number of terms × average of first and last.

  Use it when: Quick AP sums.

• Small ratio, finite total

  S∞ exists only if the ratio is between −1 and 1.

  Use it when: Infinite GP.

• Cubes are the square of the sum

  1³ + 2³ + … + n³ = (1 + 2 + … + n)².

  Use it when: Σk³.

• AM above GM

  Arithmetic mean is always at least the geometric mean; equal only for equal numbers.

  Use it when: Inequalities and least values.

Sequence and Series: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Sequence and Series

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

• ✗Mistake: Using nd instead of (n − 1)d

  Students count the first term as a step.

  Wrong working

  T₁₀ of 2, 5, 8, … = 2 + 10 × 3 = 32.

  Right working

  2 + 9 × 3 = 29.

  ✓Do this: Check your formula with n = 1: it must give a.

  Usually costs: 1 mark, and every later step.

• ✗Mistake: Applying S∞ when |r| ≥ 1

  Students plug into a/(1 − r) without checking.

  Wrong working

  1 + 2 + 4 + … = 1/(1 − 2) = −1.

  Right working

  r = 2, so no sum to infinity exists.

  ✓Do this: Write '|r| < 1, so S∞ exists' before using it.

  Usually costs: Full marks for the part.

• ✗Mistake: Mixing up Σk² and (Σk)²

  The formulas look similar.

  Wrong working

  1² + 2² + 3² = (6)² = 36.

  Right working

  1 + 4 + 9 = 14 = 3 × 4 × 7/6. It is Σk³ that equals (Σk)².

  ✓Do this: Test each formula with n = 2.

  Usually costs: 2-3 marks.

• ✗Mistake: Forgetting the constant term in a split sum

  Σ1 from 1 to n is n, not 1.

  Wrong working

  Σ(k² + k + 1) = Σk² + Σk + 1.

  Right working

  = Σk² + Σk + n.

  ✓Do this: Write Σ1 = n explicitly.

  Usually costs: 1 mark.

• ✗Mistake: Wrong number of terms when counting multiples

  Students divide the range by d without adding 1.

  Wrong working

  Two-digit multiples of 7: (98 − 14)/7 = 12 terms.

  Right working

  (98 − 14)/7 + 1 = 13 terms.

  ✓Do this: Use l = a + (n − 1)d to find n.

  Usually costs: 1-2 marks.

Practice questions on Sequence and Series

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

1. Q1 · 1 mark · MCQExam level

   The common ratio of the GP 2, −6, 18, … is:
   (a) 3
   (b) −3
   (c) −1/3
   (d) 4

   Hint

   T₂/T₁.

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   The arithmetic mean of 7 and 15 is:
   (a) 8
   (b) 10
   (c) 11
   (d) 22

   Hint

   (a + b)/2.

   Answer

   (c)

3. Q3 · 1 mark · Very short answerExam level

   Find 1 + 2 + 3 + … + 10.

   Hint

   n(n + 1)/2.

   Answer

   55

4. Q4 · 1 mark · Very short answerExam level

   Find the geometric mean of 4 and 25.

   Hint

   √(ab).

   Answer

   10

5. Q5 · 2 marks · Short answerExam level

   Find the sum to infinity of 6 + 2 + 2/3 + ….

   Hint

   r = 1/3.

   Answer

   6/(1 − 1/3) = 9

6. Q6 · 2 marks · Short answerHarder exam level

   Insert 3 arithmetic means between 5 and 25.

   Hint

   d = (25 − 5)/4.

   Answer

   d = 5: 10, 15, 20

7. Q7 · 2 marks · Short answerHarder exam level

   Find the sum of the first 10 terms of 1 + 3 + 9 + ….

   Hint

   a = 1, r = 3.

   Answer

   (3¹⁰ − 1)/2 = 59 048/2 = 29 524

8. Q8 · 3 marks · Short answerHarder exam level

   Four numbers in AP have sum 32 and the sum of their squares is 276. Find them.

   Hint

   a − 3d, a − d, a + d, a + 3d.

   Answer

   4a = 32 ⇒ a = 8. 4a² + 20d² = 276 ⇒ 20d² = 20 ⇒ d = ±1. Numbers 5, 7, 9, 11.

9. Q9 · 2 marks · Short answerHarder exam level

   Insert 2 geometric means between 3 and 192.

   Hint

   3r³ = 192.

   Answer

   r³ = 64, r = 4: 12, 48

10. Q10 · 3 marks · Short answerHarder exam level

    Find (i) 1³ + 2³ + … + 15³ (ii) 11² + 12² + … + 20².

    Hint

    For (ii), subtract Σ up to 10 from Σ up to 20.

    Answer

    (i) (15 × 16/2)² = 120² = 14 400. (ii) 20 × 21 × 41/6 − 10 × 11 × 21/6 = 2870 − 385 = 2485.

11. Q11 · 3 marks · Short answerHarder exam level

    The 3rd term of a GP is 12 and the 6th term is 96. Find the sum of its first 8 terms.

    Hint

    Divide T₆ by T₃.

    Answer

    r³ = 8 ⇒ r = 2; a = 12/4 = 3; S₈ = 3(2⁸ − 1) = 765

12. Q12 · 3 marks · Short answerHarder exam level

    Find the nth term and the sum of n terms of 2 + 5 + 10 + 17 + 26 + ….

    Hint

    Differences 3, 5, 7, 9, ….

    Answer

    Tₙ = 2 + ((n − 1)/2)[6 + 2(n − 2)] = 2 + (n − 1)(n + 1) = n² + 1. Sₙ = n(n + 1)(2n + 1)/6 + n.

13. Q13 · 5 marks · Long answerHarder exam level

    (a) The sum of the first n terms of an AP is 2n² + 3n. Find its nth term. [2]
    (b) Find the first term and common difference. [1]
    (c) Which term of this AP is 85? [2]

    Hint

    Tₙ = Sₙ − Sₙ₋₁.

    Answer

    (a) Tₙ = 2n² + 3n − [2(n − 1)² + 3(n − 1)] = 4n + 1. (b) a = 5, d = 4. (c) 4n + 1 = 85 ⇒ n = 21.

14. Q14 · 3 marks · Short answerHOTS / competency

    The sum of an infinite GP is 15 and the sum of the squares of its terms is 45. Find the GP.

    Hint

    The squares form a GP with ratio r².

    Answer

    a/(1 − r) = 15 and a²/(1 − r²) = 45. Dividing: a/(1 + r) = 3. So 15(1 − r) = 3(1 + r) ⇒ r = 2/3, a = 5. GP: 5, 10/3, 20/9, …

15. Q15 · 3 marks · Short answerHOTS / competency

    Find the least value of 4ˣ + 4¹⁻ˣ for real x, and the value of x at which it occurs.

    Hint

    The product of the two terms is constant.

    Answer

    Both terms are positive with product 4. AM ≥ GM: 4ˣ + 4¹⁻ˣ ≥ 2√4 = 4, with equality when 4ˣ = 4¹⁻ˣ, i.e. x = 1/2. Least value 4.

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. 1.nth term of an AP?Show answer

   a + (n − 1)d.

2. 2.Sum of n terms of an AP?Show answer

   (n/2)[2a + (n − 1)d] = (n/2)(a + l).

3. 3.Sum of n terms of a GP (r > 1)?Show answer

   a(rⁿ − 1)/(r − 1).

4. 4.Sum to infinity of a GP?Show answer

   a/(1 − r), only if |r| < 1.

5. 5.Three terms in GP for word problems?Show answer

   a/r, a, ar.

6. 6.Two numbers from AM A and GM G?Show answer

   A ± √(A² − G²).

7. 7.Σk²?Show answer

   n(n + 1)(2n + 1)/6.

8. 8.Σk³?Show answer

   [n(n + 1)/2]².

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you find which term of an AP equals a given number?
• ☐Can you sum all multiples of 7 between two limits?
• ☐Can you insert 4 AMs or 3 GMs between two numbers?
• ☐Can you find n for a GP with a given sum?
• ☐Can you decide whether S∞ exists?
• ☐Can you prove AM ≥ GM and use it to find a least value?
• ☐Can you sum 1·3 + 2·4 + … to n terms?
• ☐Can you find Tₙ of 3, 7, 13, 21, … by differences?

Sequence and Series: frequently asked questions