Latest
  • Admissions openClass 11 Science, 2027‑28: JEE, NEET and MHT‑CET with junior college and hostel under one roofApply now
  • IMGSAT 2027Free scholarship and admission test for Class 10 students, every Saturday and Sunday at our Nashik campusRegister
  • Free Foundation 2026‑27Evening classes for Class 10 in Physics, Chemistry, Maths and Biology, taught by our IITian and doctor facultyJoin free

+91 70303 00666

Maths · Class 11 · Chapter 8

Sequences and Series

You met arithmetic progressions in Class 10. This chapter adds geometric progressions and the inequality between the arithmetic and geometric means, which turns out to be one of the quickest tools for maximum and minimum problems in JEE.

In this chapter: sequences as functions on N, series and sigma notation, a quick review of AP, the nth term and sum of a GP, sum of an infinite GP, arithmetic and geometric means, the AM ≥ GM inequality, and the sums of n, n² and n³.

Sequences and series

A sequence is an ordered list of numbers a1, a2, a3, ..., where an is the nth term. Formally it is a function whose domain is N (an infinite sequence) or {1, 2, ..., k} (a finite sequence). A sequence may be given by a formula, like an = n² + 1, or by a recurrence, like the Fibonacci sequence: a1 = a2 = 1 and an = an−1 + an−2 for n > 2, giving 1, 1, 2, 3, 5, 8, ....

Adding the terms gives a series: a1 + a2 + ... + an, written compactly as ∑ak (k from 1 to n). We write Sn for the sum of the first n terms. A sequence whose terms follow a definite pattern is called a progression.

Arithmetic progression (review)

In an AP each term exceeds the previous one by a fixed common difference d.

an = a + (n − 1)dSn = (n/2)[2a + (n − 1)d] = (n/2)(a + l)l = last term

The arithmetic mean of a and b is A = (a + b)/2, and a, A, b form an AP. To insert n numbers between a and b so that the whole list is an AP, use d = (b − a)/(n + 1). Adding a constant to every term of an AP, or multiplying every term by a non-zero constant, gives another AP.

Geometric progression

A sequence of non-zero numbers is a GP if the ratio of each term to the previous one is a fixed number r, the common ratio (r ≠ 0). With first term a:

a, ar, ar², ar³, ...an = arn−1Sn = a(rn − 1)/(r − 1) = a(1 − rn)/(1 − r)r ≠ 1Sn = nar = 1

Use the first form of Sn when r > 1 and the second when r < 1; both are the same formula, but choosing well keeps the numbers positive. For example, 1 + 3 + 9 + ... to 8 terms = (38 − 1)/(3 − 1) = 6560/2 = 3280.

Worked example: Which term of the GP 2, 2√2, 4, ... is 128?
Solution: a = 2 and r = 2√2/2 = √2. Set 2(√2)n−1 = 128, so (√2)n−1 = 64 = 26 = (√2)12. Hence n − 1 = 12 and 128 is the 13th term.

Sum of an infinite GP

If |r| < 1, then rn gets closer and closer to 0 as n grows, and Sn approaches a fixed value:

S∞ = a/(1 − r)only when −1 < r < 1

So 1 + 1/2 + 1/4 + ... = 1/(1 − 1/2) = 2. If |r| ≥ 1 the infinite series has no finite sum. A recurring decimal is a neat application: 0.333... = 3/10 + 3/100 + ... = (3/10)/(1 − 1/10) = 1/3.

Geometric mean and inserting GMs

The geometric mean of two positive numbers a and b is G = √(ab); then a, G, b form a GP. To insert n numbers between a and b so that the whole list is a GP, the common ratio is

r = (b/a)1/(n+1)

and the inserted numbers are ar, ar², ..., arn. Inserting 3 GMs between 1 and 256 needs r4 = 256, so r = 4 and the means are 4, 16 and 64. (Taking r = −4 also satisfies r4 = 256; unless the question says otherwise, take the real positive ratio.)

Relation between AM and GM

For positive a and b, with A = (a + b)/2 and G = √(ab):

A − G = (a + b)/2 − √(ab) = (√a − √b)²/2 ≥ 0So A ≥ G, with equality only when a = b
AM ≥ GM seen in a semicirclewww.iitmedicoguide.comABOCDabCD = √(ab)(GM)OD = (a + b)/2(radius = AM)CD is half of a chord,so it cannot exceedthe radius OD:√(ab) ≤ (a + b)/2Equality only when Cis at O, i.e. a = b.www.iitmedicoguide.com
With AC = a and CB = b, the perpendicular CD equals √(ab) and the radius equals (a + b)/2. A half-chord is never longer than the radius, which is the AM ≥ GM inequality in picture form.

Also, if you know A and G, then a + b = 2A and ab = G², so a and b are the roots of x² − 2Ax + G² = 0.

Worked example: (a) The AM and GM of two positive numbers are 10 and 8. Find the numbers. (b) Find the minimum value of x + 9/x for x > 0.
Solution: (a) a + b = 20 and ab = 64, so they are the roots of x² − 20x + 64 = 0, that is, (x − 16)(x − 4) = 0. The numbers are 16 and 4.
(b) By AM ≥ GM on the positive numbers x and 9/x: (x + 9/x)/2 ≥ √(x · 9/x) = 3, so x + 9/x ≥ 6. Equality holds when x = 9/x, that is, x = 3. The minimum is 6.

The method in (b) works whenever the product of the terms is a constant. Check that equality can actually happen; otherwise the bound is not the minimum.

Sums of special series

1 + 2 + ... + n = n(n + 1)/21² + 2² + ... + n² = n(n + 1)(2n + 1)/61³ + 2³ + ... + n³ = [n(n + 1)/2]²

To sum a series whose nth term is a polynomial in n, expand an and apply these term by term. For example, if an = n(n + 1) = n² + n, then Sn = n(n + 1)(2n + 1)/6 + n(n + 1)/2 = n(n + 1)(n + 2)/3.

Common mistakes: (1) Using a/(1 − r) when |r| ≥ 1. (2) Writing the nth term of a GP as arn; it is arn−1. (3) Using Sn = a(rn − 1)/(r − 1) when r = 1, where the formula breaks down. (4) Applying AM ≥ GM to numbers that are not all positive. (5) Taking the lower bound from AM ≥ GM as the minimum without checking that equality is possible. (6) Dividing by n instead of n + 1 when inserting n means.

Exam focus

  • nth term and sum of AP and GP, including finding which term equals a given value.
  • Sum of an infinite GP and recurring decimals as fractions.
  • Inserting arithmetic and geometric means between two numbers.
  • AM ≥ GM for finding minimum or maximum values of expressions.
  • Finding two numbers from their AM and GM.
  • Sums of n, n² and n³ for series with a polynomial nth term.

Practice questions

The 20th term of the AP 3, 7, 11, ... is:

  1. 79
  2. 83
  3. 80
  4. 76
Show answer
A. 3 + 19 × 4 = 79.

The sum of the first 10 terms of the GP 1, 2, 4, 8, ... is:

  1. 1024
  2. 1023
  3. 2047
  4. 512
Show answer
B. (210 − 1)/(2 − 1) = 1023.

Which term of the GP 2, 2√2, 4, ... is 128?

  1. 12th
  2. 14th
  3. 13th
  4. 7th
Show answer
C. (√2)n−1 = 64 = (√2)12, so n = 13.

The sum of the infinite GP 1 − 1/3 + 1/9 − 1/27 + ... is:

  1. 3/2
  2. 2/3
  3. 1/2
  4. 3/4
Show answer
D. a = 1, r = −1/3: 1/(1 + 1/3) = 3/4.

The AM and GM of two positive numbers are 10 and 8. The numbers are:

  1. 12 and 8
  2. 16 and 4
  3. 18 and 2
  4. 15 and 5
Show answer
B. Sum 20 and product 64: only 16 and 4 fit.

For x > 0, the minimum value of x + 9/x is:

  1. 9
  2. 3
  3. 6
  4. 10
Show answer
C. AM ≥ GM gives x + 9/x ≥ 2√9 = 6, reached at x = 3.

Three geometric means with positive common ratio are inserted between 1 and 256. The third of them is:

  1. 16
  2. 32
  3. 128
  4. 64
Show answer
D. r4 = 256 gives r = 4; the means are 4, 16, 64.

The value of 1² + 2² + ... + 10² is:

  1. 385
  2. 55
  3. 3025
  4. 285
Show answer
A. 10 × 11 × 21/6 = 385.
Call WhatsApp Apply
Chat with us on WhatsApp