Sequence & Series

Arithmetic & Geometric Progressions, Special Series (Applied Math)

1. Introduction

Sequence

A sequence is an arrangement of numbers in a definite order according to some rule (

a_1, a_2, ..., a_n

Fibonacci: A sequence where

a_n = a_{n-1} + a_{n-2}

for

n > 2

, with

a_1=1, a_2=1

. (1, 1, 2, 3, 5...)

Series

The sum of terms of a sequence is called a series (

a_1 + a_2 + ... + a_n

Represented by

\sum

(Sigma)- '∑' notation.

2. Arithmetic Progression (A.P.)

A sequence where the difference between consecutive terms is constant (

d = a_{n} - a_{n-1}

General Term

a_n = a + (n - 1)d

Sum of n Terms

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

S_n = \frac{n}{2}(a + l)

ℹ️

Arithmetic Mean

Arithmetic Mean (AM) between

a

and

b

A = \frac{a+b}{2}

⚠️

Key Property

n^{th}

term from end =

l - (n-1)d

, where

l

is the last term.

3. Geometric Progression (G.P.)

A sequence where the ratio of consecutive terms is constant (

r = \frac{a_n}{a_{n-1}}

General Term

a_n = a \cdot r^{n-1}

Sum (Finite)

S_n = \frac{a(r^n - 1)}{r - 1}

r > 1

, or

S_n = \frac{a(1 - r^n)}{1 - r}

r < 1

Infinite GP Sum

S_\infty = \frac{a}{1 - r}

, provided

|r \mid < 1

ℹ️

Geometric Mean

Geometric Mean (GM) between

a

and

b

G = \sqrt{ab}

4. Relationships & Applications

AM vs GM Inequality

AM \ge GM

. For any distinct positive real numbers

a, b

\frac{a+b}{2} > \sqrt{ab}

Used for finding Min/Max values.

Real World Applications

💡

Compound Interest

Compound Interest growth is a Geometric Progression where

A = P(1 + \frac{r}{100})^n

⚠️

Depreciation

Depreciation (

V = P(1 - \frac{r}{100})^n

) also follows a GP model with

r < 1

Linear (AP) vs Exponential (GP) Growth

Comparing arithmetic growth ($a+d$) vs geometric growth ($a \cdot r^n$).
Notice how GP (Exponential) overtakes AP (Linear) rapidly.

5. Sum of Special Series

Sum of Natural Numbers

\sum_{k=1}^n k = \frac{n(n+1)}{2}

Sum of Squares

\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}

Sum of Cubes

\sum_{k=1}^n k^3 = [\frac{n(n+1)}{2}]^2

📝 Practice

Concept Checks

Find the 10th term of the sequence defined by

a_n = n(n-2) / 2

✓ Show Solution

a_{10} = 10(10-2) / 2 = 10(8)/2 = 80/2 = 40

EASY

How many terms are in the AP: 7, 13, 19, ..., 205?

✓ Show Solution

a = 7, d = 6, a_n = 205

205 = 7 + (n-1)6

198 = (n-1)6

33 = n-1 \Rightarrow n = 34

MEDIUM

Find the sum of odd integers from 1 to 2001.

✓ Show Solution

AP: 1, 3, 5, ..., 2001.

a=1, d=2, l=2001

Find n:

2001 = 1 + (n-1)2 \Rightarrow 2000 = 2(n-1) \Rightarrow n=1001

Sum =

\frac{n}{2}(a+l) = \frac{1001}{2}(1 + 2001) = \frac{1001}{2}(2002) = 1001 \times 1001 = 1002001

MEDIUM

Find the sum of the infinite GP:

5, \frac{20}{7}, \frac{80}{49}, ...

✓ Show Solution

a = 5

. Ratio

r = \frac{20/7}{5} = \frac{4}{7}

|r \mid < 1

? Yes,

4/7 < 1

S_\infty = \frac{a}{1-r} = \frac{5}{1 - 4/7} = \frac{5}{3/7} = \frac{35}{3}

HARD

If AM and GM of two numbers are 10 and 8, find the numbers.

✓ Show Solution

AM =

\frac{a+b}{2} = 10 \Rightarrow a+b = 20

. GM =

\sqrt{ab} = 8 \Rightarrow ab = 64

Quadratic eq:

x^2 - (a+b)x + ab = 0 \Rightarrow x^2 - 20x + 64 = 0

(x-16)(x-4) = 0

. Numbers are 4 and 16.

MEDIUM

A machine costs ₹50,000. It depreciates by 10% each year. Find its value after 3 years.

✓ Show Solution

Model as GP (

r = 1 - 0.10 = 0.90

V = P(1-i)^n = 50000(0.9)^3

= 50000 \times 0.729 = 36,450

Value is ₹36,450.

MEDIUM

🔥 CBSE Exam Priority Questions

Most Likely Questions

The sum of first 3 terms of a GP is 13/12 and their product is -1. Find the terms.

✓ Show Solution

Let terms be

\frac{a}{r}, a, ar

Product

\frac{a}{r} \cdot a \cdot ar = a^3 = -1 \Rightarrow a = -1

Sum:

-\frac{1}{r} - 1 - r = \frac{13}{12}

. Solve quadratic for

r

Results in sequences like

-4/3, -1, -3/4

-3/4, -1, -4/3

HARD

Find the sum to n terms:

8 + 88 + 888 + ...

✓ Show Solution

S = 8(1 + 11 + 111 + ...)

= \frac{8}{9}(9 + 99 + 999 + ...)

= \frac{8}{9}((10-1) + (100-1) + ...)

= \frac{8}{9}[\frac{10(10^n - 1)}{10-1} - n]

HARD

Insert 3 numbers between 1 and 256 so that the resulting sequence is a GP.

✓ Show Solution

1, G_1, G_2, G_3, 256

. Total 5 terms.

a_5 = ar^4 = 256 \Rightarrow 1 \cdot r^4 = 256 \Rightarrow r = 4

Numbers:

1 \times 4 = 4

4 \times 4 = 16

16 \times 4 = 64

MEDIUM

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1. Introduction

Sequence

A sequence is an arrangement of numbers in a definite order according to some rule (

a_1, a_2, ..., a_n

Fibonacci: A sequence where

a_n = a_{n-1} + a_{n-2}

for

n > 2

, with

a_1=1, a_2=1

. (1, 1, 2, 3, 5...)

Series

The sum of terms of a sequence is called a series (

a_1 + a_2 + ... + a_n

Represented by

\sum

(Sigma)- '∑' notation.

2. Arithmetic Progression (A.P.)

A sequence where the difference between consecutive terms is constant (

d = a_{n} - a_{n-1}

General Term

a_n = a + (n - 1)d

Sum of n Terms

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

S_n = \frac{n}{2}(a + l)

ℹ️

Arithmetic Mean

Arithmetic Mean (AM) between

a

and

b

A = \frac{a+b}{2}

⚠️

Key Property

n^{th}

term from end =

l - (n-1)d

, where

l

is the last term.

3. Geometric Progression (G.P.)

A sequence where the ratio of consecutive terms is constant (

r = \frac{a_n}{a_{n-1}}

General Term

a_n = a \cdot r^{n-1}

Sum (Finite)

S_n = \frac{a(r^n - 1)}{r - 1}

r > 1

, or

S_n = \frac{a(1 - r^n)}{1 - r}

r < 1

Infinite GP Sum

S_\infty = \frac{a}{1 - r}

, provided

|r \mid < 1

ℹ️

Geometric Mean

Geometric Mean (GM) between

a

and

b

G = \sqrt{ab}

4. Relationships & Applications

AM vs GM Inequality

AM \ge GM

. For any distinct positive real numbers

a, b

\frac{a+b}{2} > \sqrt{ab}

Used for finding Min/Max values.

Real World Applications

💡

Compound Interest

Compound Interest growth is a Geometric Progression where

A = P(1 + \frac{r}{100})^n

⚠️

Depreciation

Depreciation (

V = P(1 - \frac{r}{100})^n

) also follows a GP model with

r < 1

Linear (AP) vs Exponential (GP) Growth

Comparing arithmetic growth ($a+d$) vs geometric growth ($a \cdot r^n$).
Notice how GP (Exponential) overtakes AP (Linear) rapidly.

5. Sum of Special Series

Sum of Natural Numbers

\sum_{k=1}^n k = \frac{n(n+1)}{2}

Sum of Squares

\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}

Sum of Cubes

\sum_{k=1}^n k^3 = [\frac{n(n+1)}{2}]^2

📝 Practice

Concept Checks

Find the 10th term of the sequence defined by

a_n = n(n-2) / 2

✓ Show Solution

a_{10} = 10(10-2) / 2 = 10(8)/2 = 80/2 = 40

EASY

How many terms are in the AP: 7, 13, 19, ..., 205?

✓ Show Solution

a = 7, d = 6, a_n = 205

205 = 7 + (n-1)6

198 = (n-1)6

33 = n-1 \Rightarrow n = 34

MEDIUM

Find the sum of odd integers from 1 to 2001.

✓ Show Solution

AP: 1, 3, 5, ..., 2001.

a=1, d=2, l=2001

Find n:

2001 = 1 + (n-1)2 \Rightarrow 2000 = 2(n-1) \Rightarrow n=1001

Sum =

\frac{n}{2}(a+l) = \frac{1001}{2}(1 + 2001) = \frac{1001}{2}(2002) = 1001 \times 1001 = 1002001

MEDIUM

Find the sum of the infinite GP:

5, \frac{20}{7}, \frac{80}{49}, ...

✓ Show Solution

a = 5

. Ratio

r = \frac{20/7}{5} = \frac{4}{7}

|r \mid < 1

? Yes,

4/7 < 1

S_\infty = \frac{a}{1-r} = \frac{5}{1 - 4/7} = \frac{5}{3/7} = \frac{35}{3}

HARD

If AM and GM of two numbers are 10 and 8, find the numbers.

✓ Show Solution

AM =

\frac{a+b}{2} = 10 \Rightarrow a+b = 20

. GM =

\sqrt{ab} = 8 \Rightarrow ab = 64

Quadratic eq:

x^2 - (a+b)x + ab = 0 \Rightarrow x^2 - 20x + 64 = 0

(x-16)(x-4) = 0

. Numbers are 4 and 16.

MEDIUM

A machine costs ₹50,000. It depreciates by 10% each year. Find its value after 3 years.

✓ Show Solution

Model as GP (

r = 1 - 0.10 = 0.90

V = P(1-i)^n = 50000(0.9)^3

= 50000 \times 0.729 = 36,450

Value is ₹36,450.

MEDIUM

🔥 CBSE Exam Priority Questions

Most Likely Questions

The sum of first 3 terms of a GP is 13/12 and their product is -1. Find the terms.

✓ Show Solution

Let terms be

\frac{a}{r}, a, ar

Product

\frac{a}{r} \cdot a \cdot ar = a^3 = -1 \Rightarrow a = -1

Sum:

-\frac{1}{r} - 1 - r = \frac{13}{12}

. Solve quadratic for

r

Results in sequences like

-4/3, -1, -3/4

-3/4, -1, -4/3

HARD

Find the sum to n terms:

8 + 88 + 888 + ...

✓ Show Solution

S = 8(1 + 11 + 111 + ...)

= \frac{8}{9}(9 + 99 + 999 + ...)

= \frac{8}{9}((10-1) + (100-1) + ...)

= \frac{8}{9}[\frac{10(10^n - 1)}{10-1} - n]

HARD

Insert 3 numbers between 1 and 256 so that the resulting sequence is a GP.

✓ Show Solution

1, G_1, G_2, G_3, 256

. Total 5 terms.

a_5 = ar^4 = 256 \Rightarrow 1 \cdot r^4 = 256 \Rightarrow r = 4

Numbers:

1 \times 4 = 4

4 \times 4 = 16

16 \times 4 = 64

MEDIUM