basic numeracy-sequences-and-series

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Basic Numeracy

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Sequences & Series

Arithmetic Progression (AP)

An arithmetic progression is a sequence in which terms increase or decrease by

a constant number called the common difference.

(i) The sequence 2, 6, 10, 14, 18, 22… is an arithmetic progression whose

first term is 2 and common difference 4.

(ii) The sequence 2, 5/2, 3 ,7/2 ,4 …is an arithmetic progression whose first

term is 2 and common difference ½.

An arithmetic progression is represented by a,(a + d), (a + 2d), (a + 3d) a + (n –

1)d

Here, a = first term

d = common difference

n = number of terms in the progression

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• The general term of an arithmetic progression is given by Tn = a + (n - 1) d.

• The sum of n terms of an arithmetic progression is given by S, = n/2 [2a + (n

– 1) d] or Sn = 2 [a + l]

where l is the last term of arithmetic progression.

• If three numbers are in arithmetic progression, the middle number is

called the arithmetic mean of the other two terms.

• If a, b, c are in arithmetic progression, then b = a+c/2

where b is the arithmetic mean.

• Similarly, if ‘n’ terms al, a2, a3… an are in AP, then the arithmetic mean of

these ‘n’ terms is given by

AM =

• If the same quantity is added or multiplied to each term of an AP, then the

resulting series is also an AP.

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• If three terms are in AP, then they can be taken as (a – d), a, (a + d).

• If four terms are in AP, then they can be taken as (a – 3d), (a – d), (a +

d), (a + 3d).

• If five terms are in AP, then they can be taken as (a – 2d), (a – d), a, (a +

d), (a + 2d).

Geometric Progression (GP)

A geometric progression is a sequence in which terms increase or decrease by a

constant ratio called the common ratio.

(i) The sequence 1, 3, 9, 27, 81…is a geometric progression whose first

term is 1 and common ratio 3.

(ii) The sequence is a geometric progression whose first

term is 1 and common ratio 1/3.

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A geometric progression is represented by a, ar, ar2…arn–1.

Here, a = first term

r = common ratio

n = number of terms in the progression.

• The general term of a geometric progression is given by Tn = an–1

• The sum to n terms of a geometric progression is given by

when r < 1

when r > 1

• If three numbers are in geometric progression, the middle number is

called the geometric mean of the other two terms.

• If a, b, c are in geometric progression, then where b is the

geometric mean.

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• Similarly, if n terms a1, a2, a3, a4,…an are in geometric progression, then

the geometric mean of 1 these n terms is given by

GM =

• For a decreasing geometric progression the sum to infinite number of

terms is

where a = first term and | r | < 1.

• If every term of a GP is multiplied by a fixed real number, then the

resulting series is also a GP.

• If every term of a GP is raised to the same power, then the resulting

series is also a GP.

• The reciprocals of the terms of a GP is also a GP.

• If three numbers are in GP, then they can be taken as a/r , a, ar.

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• If four numbers are in GP, then they can be taken as

• If five numbers are in GP, then they can be taken as

Harmonic Progression (HP)

If the reciprocals of the terms of a series form an arithmetic progression, then the

series is called a harmonic progression.

(i) The sequence 4/3, 3/2, 12/7, …. is a harmonic progression as 3/4, 2/3,

7/12 is in arithmetic progression.

• If a, b, c are in harmonic progression, then b = 2ac / a+c where b is the

harmonic mean.

Sum of Natural Series

• The sum of the first n natural numbers = n (n+1) / 2 .

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• The sum of the square of the first n natural numbers = n (n+1) (2n+1) / 6

• The sum of the cubes of the first n natural numbers =

• The sum of first n even numbers = n(n + 1)

• The sum of first n odd numbers = n2

Example 1: Find the nth term and the fifteenth term of the arithmetic

progression 3, 9, 15, 21…

Solution.

In the given AP we have a = 3, d = (9 – 3) = 6

Tn = a + (n – 1)d = 3 + (n – 1)6 = 6n – 3

T15 = (6 × 15 – 3) = 87

Example 2: Find the 10th term of the AP 13, 8, 3, –2,…

Solution.

In the given AP, we have a = 13, d = (8 –13) = –5

Tn = a + (n – 1)d = 13 + (n – 1)(–5) = 18 – 5n

T10 = 18 – 5 (10) = –32

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Example 3: The first term of an AP is -1 and the common difference is -3, the

12th term is

Solution.

T1 = a = –1, d = –3

Tn = a + (n – 1)d = –1 + (n – 1)(–3) = 2 – 3n

T12 = 2 – 3 × 12 = –34

Example 4: Which term of the AP 10, 8, 6, 4… is –28?

Solution.

We have, a = 10,d = (8 – 10) = –2, Tn = –28

Tn = a + (n – 1)d – 28 = 10 + (n – 1)(–2) = n = 20

Example 5: The 8th term of an AP is 17 and the 19th term is 39. Find the

20th term.

Solution.

T8 = a + 7d =17 ...(i)

T19 = a + 18d = 39 ... (ii)

On subtracting Eq. (i) from Eq. (ii), we get 11d = 22

= d = 2

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Putting d = 2 in Eq. (i),

we get

a + 7(2) = 17

a = (17 – 14) = 3

. . . First term = 3, Common difference = 2

T20= a + 19d

= 3 + 19(2)

= 41

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