Sequences and series: Class 11 Mathematics Notes, Short Questions and Numericals

Sequences and series, arithmetic and geometric progressions, nth terms and sums, means, harmonic progressions, relation between A.M., G.M. and H.M., and the sum to infinity of a geometric series.

Key points

A sequence is a list of numbers in a definite order, such as 2, 5, 8, 11. Each number is a term. The nth term is written aₙ. A series is the sum of the terms of a sequence, so 2 + 5 + 8 + 11 is a series. A sequence is finite if it ends and infinite if it goes on without end.

An arithmetic progression (A.P.) is a sequence in which each term is found by adding a fixed number, the common difference d, to the term before it. If the first term is a, the terms are a, a + d, a + 2d, and so on. The nth term is aₙ = a + (n − 1)d. For example, 5, 8, 11, 14 has a = 5 and d = 3.

The sum of the first n terms of an A.P. is Sₙ = (n/2)[2a + (n − 1)d]. It can also be written Sₙ = (n/2)(a + l), where l is the last term. The second form is quick when the first and last terms are known. The sum of the first n natural numbers is n(n + 1)/2.

The arithmetic mean (A.M.) of two numbers a and b is (a + b)/2. It is the middle term of the A.P. a, A, b. To insert n arithmetic means between a and b, we form an A.P. with n + 2 terms whose first term is a and last term is b, and common difference d = (b − a)/(n + 1).

A geometric progression (G.P.) is a sequence in which each term is found by multiplying the term before it by a fixed number, the common ratio r. The terms are a, ar, ar², and so on. The nth term is aₙ = arⁿ⁻¹. The ratio r is found by dividing any term by the one before it. For 3, 6, 12, 24 we have a = 3 and r = 2.

Sum of a G.P.:

For r ≠ 1: Sₙ = a(rⁿ − 1)/(r − 1), which is convenient when r > 1.
Also Sₙ = a(1 − rⁿ)/(1 − r), which is convenient when r < 1.
For r = 1 all terms are equal and Sₙ = na.

An infinite G.P. has a sum only when |r| < 1, because then rⁿ becomes very small as n grows. In that case S∞ = a/(1 − r). For example 8 + 4 + 2 + 1 + … has a = 8 and r = 1/2, so S∞ = 8/(1/2) = 16. This idea also converts a recurring decimal to a fraction. If |r| ≥ 1 the sum does not settle on a number.

The geometric mean (G.M.) of two positive numbers a and b is √(ab). It is the middle term of the G.P. a, G, b. If a, b and c are in G.P. then b² = ac.

A harmonic progression (H.P.) is a sequence whose reciprocals form an A.P. For example 1, 1/2, 1/3, 1/4 is an H.P. because 1, 2, 3, 4 is an A.P. The harmonic mean (H.M.) of a and b is 2ab/(a + b). To work on an H.P., convert it to an A.P. by taking reciprocals.

For two positive numbers a and b, the three means are related by G² = A × H, so the G.M. is the G.M. of the A.M. and the H.M. Also A ≥ G ≥ H, with equality only when a = b. Series with other patterns can be summed using standard results: Σn = n(n + 1)/2, Σn² = n(n + 1)(2n + 1)/6 and Σn³ = [n(n + 1)/2]².

Study tip: First decide whether the sequence has a common difference or a common ratio, and write a, d or r on the first line. For word problems with unknown a and d, form two equations from the given terms and subtract them. Check the answer by listing the first few terms.

Definitions and flashcards

Say the meaning first, then open the card to check.

Sequence

A list of numbers in a definite order.

Series

The sum of the terms of a sequence.

Common difference

The fixed number added to get the next term in an A.P.

Common ratio

The fixed number multiplied to get the next term in a G.P.

Arithmetic mean

The value (a + b)/2 between a and b.

Geometric mean

The value √(ab) of two positive numbers.

Harmonic progression

A sequence whose reciprocals are in A.P.

Sum to infinity

The limit a/(1 − r) of a G.P. series when |r| < 1.

Formula sheet

Name	Formula	What it means
nth term of an A.P.	aₙ = a + (n − 1)d	First term a, common difference d.
Sum of an A.P.	Sₙ = (n/2)[2a + (n − 1)d]	Or (n/2)(a + l).
nth term of a G.P.	aₙ = arⁿ⁻¹	Common ratio r.
Sum of a G.P.	Sₙ = a(rⁿ − 1)/(r − 1)	For r ≠ 1.
Sum to infinity	S∞ = a/(1 − r)	Only for \|r\| < 1.
Means of a and b	A = (a + b)/2, G = √(ab), H = 2ab/(a + b)	G² = A × H.
Sum of squares	Σn² = n(n + 1)(2n + 1)/6	For n natural numbers.

Short questions with model answers

These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.

Define an arithmetic progression and write its nth term. (2 marks)

An A.P. is a sequence in which the difference between any term and the one before it is the same fixed number d. If the first term is a, the nth term is a + (n − 1)d.

When does an infinite G.P. have a sum? Write the formula. (2 marks)

It has a sum only if |r| < 1. Then the sum is S∞ = a/(1 − r), where a is the first term and r is the common ratio.

Find the arithmetic mean and the geometric mean of 4 and 16. (3 marks)

A.M. = (4 + 16)/2 = 10. G.M. = √(4 × 16) = √64 = 8. The G.M. is smaller than the A.M., as it should be for unequal positive numbers.

Insert three arithmetic means between 2 and 14. (3 marks)

We need an A.P. of 5 terms with first term 2 and last term 14. So 14 = 2 + 4d, which gives d = 3. The terms are 2, 5, 8, 11, 14. The three means are 5, 8 and 11.

Derive the formula for the sum of the first n terms of an A.P. (5 marks)

Let Sₙ = a + (a + d) + … + [a + (n − 1)d]. Write the same sum in reverse: Sₙ = [a + (n − 1)d] + … + (a + d) + a. Add the two lines term by term. Each pair adds to 2a + (n − 1)d, and there are n pairs. So 2Sₙ = n[2a + (n − 1)d]. Therefore Sₙ = (n/2)[2a + (n − 1)d].

Derive the sum of the first n terms of a G.P. and the sum to infinity. (5 marks)

Let Sₙ = a + ar + … + arⁿ⁻¹. Multiply by r: rSₙ = ar + ar² + … + arⁿ. Subtract: Sₙ − rSₙ = a − arⁿ. So Sₙ(1 − r) = a(1 − rⁿ), and Sₙ = a(1 − rⁿ)/(1 − r) for r ≠ 1. If |r| < 1, then rⁿ tends to 0 as n grows, so the sum to infinity is S∞ = a/(1 − r).

Numericals with worked steps

In an A.P. a = 5 and d = 3. Find the 20th term and the sum of the first 20 terms.

a₂₀ = a + 19d = 5 + 19(3) = 5 + 57 = 62.
S₂₀ = (20/2)(a + l) = 10(5 + 62).
= 10 × 67 = 670.

Answer: a₂₀ = 62 and S₂₀ = 670.

Find the 8th term and the sum of the first 8 terms of the G.P. 3, 6, 12, …

a = 3 and r = 6/3 = 2.
a₈ = ar⁷ = 3 × 128 = 384.
S₈ = a(r⁸ − 1)/(r − 1) = 3(256 − 1)/1.
= 3 × 255 = 765.

Answer: a₈ = 384 and S₈ = 765.

The 3rd term of an A.P. is 11 and the 7th term is 23. Find the first term, the common difference and the 10th term.

a + 2d = 11 and a + 6d = 23.
Subtract: 4d = 12, so d = 3.
a = 11 − 2(3) = 5.
a₁₀ = 5 + 9(3) = 32.

Answer: a = 5, d = 3 and a₁₀ = 32.

Find the sum to infinity of 12 + 6 + 3 + 1.5 + …

a = 12 and r = 6/12 = 1/2.
|r| < 1, so the sum exists.
S∞ = a/(1 − r) = 12/(1/2) = 24.

Answer: S∞ = 24.

Find A, G and H for the numbers 4 and 16 and verify G² = A × H.

A = (4 + 16)/2 = 10.
G = √(4 × 16) = 8.
H = 2(4)(16)/(4 + 16) = 128/20 = 6.4.
G² = 64 and A × H = 10 × 6.4 = 64.

Answer: A = 10, G = 8, H = 6.4, and G² = A × H = 64.

Common mistakes

Using n instead of n − 1 in the nth term formulas aₙ = a + (n − 1)d and aₙ = arⁿ⁻¹.
Applying the sum to infinity formula when |r| ≥ 1.
Forgetting that the H.P. must be turned into an A.P. by taking reciprocals before applying A.P. formulas.
Mixing up the two forms of the G.P. sum, which gives a negative answer for r > 1 if the wrong one is used.
Finding the common ratio as the difference of two terms instead of the quotient.

Practise this chapter

Practice questions and timed test Printable worksheet