X Maths Chapter 1 - Arithmetic Sequences

Arithmetic Sequence 7

Q3) Write down the sequence of natural numbers which leave

remainder 2 on division by 4 and write its algebraic form.

Ans) Sequence is

2 , 6 , 10 , 14 , . . . . . .

nth term = dn + f – d f=2, d=6−2=4

= 4n − 2 f − d = 2 − 4 =−2

d = 4, here each term of the sequence is 2 less than
the multiples of 4 . So, algebraic form is 4n − 2 .

Q4) Write down the sequence of natural numbers which leave

remainder 3 on division by 4 and write its algebraic form.

Ans) Sequence is

3 , 7 , 11 , 15 , . . . . . . . .

nth term = dn + f – d f=3, d=7−3=4

= 4n − 1 f − d = 3 − 4 =−1

d = 4, here each term of the sequence is 1 less than
the multiples of 4 . So, algebraic form is 4n − 1 .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1

Online Class – X − 09 08 / 07 / 2021

1. Arithmetic Sequences – Class 7

To view class

The algebraic form of an arithmetic sequence
is of the form a n + b , where ‘ a ’ is the common difference

Conversely
Any sequence of the form a n + b is an arithmetic sequence

xn = a n + b a=d
or b =f−d

xn = d n + (f − d)

Q1) The algebraic form of an arithmetic sequence is 5n + 3 .
a) Find the common difference .
b) Write the arithmetic sequence.

Ans)
a) Common difference = 5
First term = 5 × 1 + 3 = 8
Second term = 5 × 2 + 3 = 13
Third term = 5 × 3 + 3 = 18
∴ Arithmetic sequence is 8 , 13 , 18 , . . . . . . .

Q2) Consider the arithmetic sequence 1 , 5 , 7 , 9 ,.....
2 6 6 6

a) Write the algebraic form of the sequence .

b) Prove that the sequence contains no natural numbers .

Ans) a) 7 5 2
6 6 6
d= − =

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

f= 1 = 1x3 = 3
2 2x3 6

xn = d n + (f − d) 2
2 3 6
= 6 n +( 6 − )

= 2 n+ 1
6 6
2 n+1
= 6

Finding algebraic form without using the formula:
3 5 7 9
Sequence is 6 , 6 , 6 , 6 ,.....

Numerator is the sequence of odd numbers , its

algebraic form is 2n + 1 and the denominator is 6 .
2 n+1
∴ Algebraic form of the given sequence is 6

b)

In the given sequence the numerator of all terms is an odd

number and the denominator is an even number 6 .
Odd number
Even number cannot be a natural number.

∴ The given sequence does not contain any natural number

term .

Q3) Consider the arithmetic sequence 1 , 2 , 3 , 4 ,.....
7 7 7 7

a) Write the algebraic form of the sequence .

b) Prove that the sequence contains all natural numbers .
2 1 1
Ans) a) d = 7 − 7 = 7

f= 1
7

xn = d n + (f − d)

= 1 n +( 1 − 1 )
7 7 7
1 n
= 7 n+0 = 7

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

Finding algebraic form without using the formula:
1 2 3 4
Sequence is 7 , 7 , 7 , 7 ,.....

Numerator is the sequence of natural numbers.

Its algebraic form is ‘n’ . Denominator is 7 .
n
∴ Algebraic form of given sequence is 7

b)

In the given sequence the numerators of all terms are

natural numbers and the denominator is 7 .

When n = 7 , 14 , 21 , 28 , 35 , . . . we get all natural numbers

as terms of this sequence .

Q4) The 8th term of an arithmetic sequence is 12 and its 12th
term is 8.
a)What is the algebraic expression for this sequence?
b) Find the 20th term.

Ans) a) X8 = 12 , X12 = 8

Common difference, d= Term difference
Position difference

= 8−12 = −4 = −1
12−8 4

First term , f = X8 − 7 d = 12 − 7 × ( −1 )

= 12 + 7 = 19

Xn = d n + f − d
= −1 × n + 19 − ( −1 )
= −n + 19 + 1
= −n + 20 = 20 − n

b) 20th term , X20 = 20 − 20 =0

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 4

Q5) Consider the arithmetic sequence 4, 7, 10, . . . . . . . . . . .

a) Write the algebraic form of the sequence .

b) Prove that the squares of all terms belong to the sequence.

Ans)

a) Sequence is 4, 7, 10, . . . . . . . . . . d=7−4=3

Its algebraic form is 3n + 1

That is each term of the sequence is 1 added to a

multiple of 3

b) Square of any term of this sequence is (3n + 1)2

(3n + 1)2 = (3n)2 + 2 × 3n × 1 + 12 (a+b)2 = a2 +2ab + b2
= 9n2 + 6n + 1

= 3(3n + 2) + 1 , this is also 1 added to a

multiple of 3 .

So we can say the squares of all terms of the given
arithmetic sequence 4 , 7 , 10 , . . .belong to the sequence .

Assignment (TB.Page 25)

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1

Answers of Maths Note 07

Assignment (TB.Page 25)

Ans) Given f = 1 = 1x2 = 2 & d= 1
3 3x2 6 6

X n = dn + f – d 1
1 2 6
= 6 n+ 6 −

= n + 2− 1
6 6
n 1
= 6 + 6

= n+1
6
n+1
For 6 to be a natural number, n+1 must be a multiple of 6.

n+1 = 6 ⇒ n = 5 ∴ x 5= 5+1 = 6 =1
n+1 = 12 ⇒ n = 11 ∴ x 11 = 6 6 =2
n+1 =18 ⇒ n = 17 ∴ x17= 12 =3
11+ 1 = 6
6
= 18 etc
17+ 1 6
6

That is ,the sequence contains all natural numbers

Ans) Given f = 1 & d= 2
3 3

X n = dn + f – d

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

= 2 n+ 1 − 2
3 3 3
2 −1
= 3 n+ 3

= 2n + −1
3 3
2 n−1
= 3

For n = 2, 2 n−1 = 2 x 2−1 = 4−1 = 3 =1
3 3 3 3

For n = 5, 2 n−1 = 2 x 5−1 = 10 −1 = 9 =3
3 3 3 3

For n = 8, 2 n−1 = 2 x 8−1 = 16−1 = 15 =5 etc
3 3 3 3

So the given arithmetic sequence contains all odd numbers.

* 2n is an even number . So 2n −1 is an odd number.

In the given sequence the numerator of all terms is an odd

number and the denominator is an odd number 3 .
Odd number
Odd number cannot be an even number.

∴ The given sequence does not contain any even number.

Ans) Given f = 5 & d = 8 – 5 = 3
X n = dn + f − d
= 3n + 5 − 3
= 3n + 2

This sequence when we divide by 3 leaves remainder 2.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

We know perfect squares are 1, 4, 9, 16, . . . . . . .
When we divide a perfect squares by 3 we get remainder as
1 or 0.
Since remainder is not 2, we can say perfect squares are
not the terms of the given sequence.

Ans) Given f = 11 &d= 14 − 11
8 8 8
3
= 8

X n= d n + f − d
3 11 3
= 8 n+ 8 − 8

= 3n + 8
8 8
3n
= 8 +1

If ‘n’ is a multiple of 8 then
3x8
8 +1 =3+1=4

3 x 16 +1=6+1=7
8
3 x 24
8 + 1 = 9 + 1 =10 etc are the whole numbers.

Sequence of whole numbers of this sequence is
4, 7, 10, . . . . . . . . . .
This is an arithmetic sequence with common difference 3.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1

Online Class – X − 10 12 / 07 / 2021

1. Arithmetic Sequences – Class 8

To view class

Sums and Terms
a) Sum of consecutive natural numbers
* Sum of 3 consecutive natural numbers

1+2+3 = 6 =3×2
2+3+4 = 9 =3×3
3 + 4 + 5 = 12 = 3 × 4

using algebra:
if we take the middle number as n , we have
(n − 1) + n + (n + 1) = 3n

So,
Sum of any 3 consecutive natural numbers is
3 times the middle number

* Sum of 5 consecutive natural numbers

1 + 2 + 3 + 4 + 5 = 15 = 5 × 3
4 + 5 + 6 + 7+ 8 = 30 = 5 × 6

using algebra:
if we take the middle number as n , we have
(n – 2) + (n − 1) + n + (n + 1) + (n + 2) = 5n

So,
Sum of any 5 consecutive natural numbers is
5 times the middle number

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

* Sum of 7 consecutive natural numbers
1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 = 7 × 4

using algebra:
if we take the middle number as n , we have

(n – 3) + (n – 2) + (n − 1) + n + (n + 1) + (n + 2) + (n + 3) = 7n
So,
Sum of any 7 consecutive natural numbers is
7 times the middle number

b) Sum of consecutive even numbers
* Sum of 3 consecutive even numbers

2 + 4 + 6 = 12 = 3 × 4
8 + 10 + 12 = 30 = 3 × 10

using algebra:
if we take the middle number as x , we have

(x − 2) + x + (x + 2) = 3 x
So,

Sum of any 3 consecutive even numbers is
3 times the middle number

c) Sum of consecutive odd numbers
* Sum of 3 consecutive odd numbers

1+ 3 + 5 =9 = 3×3
5 + 7 + 9 = 21 = 3 × 7

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

using algebra:
if we take the middle number as x , we have
(x − 2) + x + (x + 2) = 3x
So,
Sum of any 3 consecutive odd numbers
is 3 times the middle number

d) Sum of consecutive multiples of 3
* Sum of 3 consecutive multiples of 3

3 + 6 + 9 = 18 = 3 × 6
9 + 12 + 15 = 36 = 3 × 12

using algebra:
if we take the middle number as x , we have

(x − 3) + x + (x + 3) = 3 x
So,

Sum of any 3 consecutive multiples of 3 is
3 times the middle number

e) Sum of consecutive terms of an arithmetic sequence
* Sum of 3 consecutive terms of an arithmetic sequence
Consider the sequence obtained by adding 1 to the
multiples of 4 .
The sequence is 5 , 9 , 13 , . . . . . . .
Considering sum of 3 consecutive terms 5, 9, 13
5 + 9 + 13 = 27 = 3 × 9
using algebra:
if we take the middle number as x , we have
(x − 4) + x + (x + 4) = 3 x

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 4

* Sum of 5 consecutive terms of an arithmetic sequence
using algebra:
if we take the middle number as x , we have
(x − 8) + (x − 4) + x + (x + 4) + (x + 8) = 5x

In general,
• Considering 3 consecutive terms of an arithmetic

sequence , if we take the common difference as ‘y’ and the
middle term as ‘x’, the terms of the sequence are
x − y, x , x + y
Sum = x − y + x + x + y = 3 x

• Considering 5 consecutive terms of an arithmetic
sequence , if we take the common difference as ‘y’ and the
middle term as ‘x’, the terms of the sequence are
x −2y , x − y , x , x + y , x + 2y
Sum = x −2y + x − y + x + x + y + x + 2y = 5 x

From above we can see,
If we consider odd number of consecutive terms of an
arithmetic sequence

Sum = Number of terms × Middle term

Middle Term = Sum

Number of terms

Activity 1
Write 4 arithmetic sequences with 500 as the sum of
the first five terms.
Ans) Given , Sum of 5 terms, S5 = 500

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 5

Number of terms = 5 ( odd number)

∴ Middle term ( 3rd term) = Sum

Number of terms

= 500 = 100
5

So sequences can be,

100, 100, 100, 100, 100 (d = 0)

98, 99, 100, 101, 102 (d = 1)

96, 98, 100, 102, 104 (d = 2)

90, 95, 100, 105, 110 (d = 5)

Activity 2

The first term of an arithmetic sequence is 10 and the

sum of the first 5 consecutive terms is 500. Write the

sequence.

Ans) Given , Sum of 5 terms, S5 = 500

Number of terms = 5 Sum
Number of terms
∴ Middle term ( 3rd term) =

= 500 = 100
5

Given , 1st term = 10

Common difference, d= Term difference
Position difference

= 100−10 = 90
3−1 2

= 45

∴ Sequence is ,

10, 55, 100, 145, 190, . . . . . . . .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 6

Activity 3

The sum of first 5 terms of an arithmetic sequence is 70. The

sum of the first 11 terms is 286.

a) What is its third term?

b) What is its sixth term?

c) What is the common difference of this sequence ?

d) What is the first term of this sequence ?

e) Write the algebraic form of this sequence .

Ans)

a) Given , S5 = 70 Su m
∴ Middle Term = X3 = No : of terms
70
= 5

= 14

b) Given , S11 = 286 Su m
No : of terms
∴ Middle Term = X6 =

= 286
11

= 26

c) Common difference, d = Term difference
Position difference
X6− X 3
d= 8− 3

= 26− 14
6−3
12
= 3 =4

d) First term, X1 = X3 – 2d

= 14 – 2 × 4

= 14 – 8

=6

e) Xn = dn + f – d
= 4n + 6 – 4

= 4n + 2

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 7

Some other peculiarities of arithmetic sequence

a) When the number of terms odd

Consider an arithmetic sequence with odd number of terms.

1, 3, 5, 7, 9, 11, 13 Here, no: of terms = 7 (odd)

We can make pair the numbers as follows

1, 3, 5, 7, 9, 11, 13 1 + 13 = 14
3 + 11 = 14
5 + 9 = 14

Sum of numbers in each pair (pair sum) = 14 In an A.S
Middle term of sequence = 7 with number
14 = 2 × 7
of terms
Pair sum = 2 × middle term odd
Pair Sum

Middle term =
2

Activity 4

Complete following

i) 4, ----, 16 Pair su m 4+ 16 20
2 2 2
Ans) Middle term = = = = 10

ii) 13,----, 35 Pair su m 13+ 35 48
2 2 2
Ans) Middle term = = = = 24

iii) 12, ----, 34
iv) 16, ----, 32

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 8

b) When the number of terms even

Consider an arithmetic sequence with even number of terms.

1, 3, 5, 7, 9, 11 Here no: of terms = 6 (even)

We can make pair of numbers as follows

1 + 11 = 12 No: of pairs = 3
3 + 9 = 12
1, 3, 5, 7, 9, 11 5 + 7 = 12

Sum of numbers in each pair (pair sum) = 12
Number of pairs = 3

Sum = 1 + 3 + 5 + 7 + 9 + 11 = 36
36 = 3 × 12
Sum = No: of pairs × Pair sum

Sum of terms = No: of pairs × Pair sum In an A.S
Pair sum = Sum of terms with number
No: of pairs
of terms
even

Activity 5 7
Find sum of following arithmetic sequences
i) 1, 2, 3, 4, 5, 6 7
Ans) No: of pairs = 3 7

Pair sum = 7 1, 2, 3, 4, 5, 6
Sum = No: of pairs × Pair sum
=3×7
= 21

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 9

ii) 1, 2, 3, 4, . . . . . . . . . . . ,50
Ans) No: of pairs = 25

Pair sum = 51
Sum = No: of pairs × Pair sum
= 25 × 51
= 1275

iii) 1, 2, 3, 4, 5, . . . . . . . . . . . . ,60
iv) 1, 2, 3, 4, 5, . . . . . . . . . . . . ,200

In algebra:
Algebraic form of any arithmetic sequence is Xn= an + b
Sequence is,
a + b, 2a + b, 3a + b, 4a + b, . . . . . . . . . . . . . . .10a + b

1st term + 10th term = (a + b) + (10a + b) = 11a + 2b

2nd term + 9th term = (2a + b) + (9a + b) = 11a + 2b

3rd term + 8th term = (3a + b) + (8a + b) = 11a + 2b

4th term + 7th term = (4a + b) + (7a + b) = 11a + 2b

5th term + 6th term = (5a + b) + (6a + b) = 11a + 2b

Here,

Sums of position of pairs =11 1+10 =2+9 =3+8

Sums of pairs of terms = 11a + 2b =4+7 = 5+6 =11

So we can conclude

In an arithmetic sequence, if the sums of positions of two
pairs of terms are equal, then the sums of the pairs of the

terms are also equal

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 10

Activity 6

The sum of fifth term and sixteenth term of an arithmetic
sequence is 50 . Find the sum of first 20 terms ?

Ans) 10 pairs
Given, 5th term + 16th term = 50
We have to find the sum of first 20 terms . X1 , X20
By pairing the 20 terms we get 10 pairs with
equal sums and with equal sums of position . X2 , X19
X3 , X18
∴ Sum of first 20 terms = No: of pairs × Pair sum X4 , X17
= 10 × 50 X5 , X16
= 500 .......

.......

Activity 7

The sum of first 5 terms of an arithmetic sequence is 100 and

the sum of its first 10 terms is 250. Find the sequence .

Ans)

Let the first five terms be X1 , X2 , X3 , X4 , X5

Sum of first 5 terms = X1 + X2 + X3 + X4 + X5 = 100
Su m of terms
∴ Middle term (3rd term (X3) ) = No : of terms

= 100 = 20
5

Given, 5 pairs
Sum of first ten terms = 250 X1 , X10

By pairing the 10 terms we get 5 pairs with X2 , X9
equal sums and with equal sums of position . X3 , X8
X4 , X7
X1, X2, X3, X4, X5, X6, X7, X8, X9, X10 X5 , X6

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 11

Sum = No: of pairs × pair sum

250 = 5 × (3rd term + 8th term)
250
3rd term + 8th term = 5

= 50

∴ 8th term (X8 ) = 50 − 3rd term

= 50 – 20

= 30

Common difference, d = Term difference
Position difference
X8− X 3
d= 8− 3

= 30− 20
8−3
10
= 5

=2

First term, X1 = X3 – 2d 3rd term, X3 = 20

= 20 – 2 × 2

= 20 – 4

= 16

∴ Sequence is 16,18, 20, 22, 24, 26, 28 , 30, 32, 34, 36, . . . . . . . .

Assignment
T.B Page 28

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 12

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1

Answers of Maths Note 08

Assignment (TB.Page 25)

Ans) Given , No of terms = 5 , sum = 30 Sum

Middle term ( 3rd term) = Number of terms

= 30 =6
5

So arithmetic sequences are

4, 5, 6, 7, 8, . . . . . . . . . . (d =1)

2, 4, 6, 8, 10, . . . . . . . . . (d =2)

0, 3, 6, 9, 12,. . . . . . . . . . (d =3)

Ans)

Given X1 + X2 + X3+ X4 = 100 Su m of terms
No : of pairs
Pair sum = X1 + X4 = X2 + X3 =

= 100 = 50
2

Given, X1 = 1, X1 + X4 = 50

1 + X4 = 50

X4 = 50 – 1 = 49

d= Term difference 1st term =1
Position difference 4th term =49
49− 1 48
∴ d= 4−1 = 3 = 16

∴ Arithmetic sequence is 1, 17, 33, 49 . . . . . . . . . . . .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

Ans)

Let the first term of the arithmetic sequence be ‘f ’ and

the common difference be ‘d ’ .

Four consecutive terms of the sequence are

f, f + d, f + 2d, f +3d

1st term + 4th term = f + (f + 3d) = 2f + 3d (1)

2nd term + 3rd term = (f + d) + (f + 2d) = 2f + 3d (2)

From equations (1) & (2)

1st term + 4th term = 2nd term + 3rd term

Ans)

Given X1 + X2 + X3+ X4 = 100 Su m of terms
Pair sum = X1 + X4 = X2 + X3 =
No : of pairs
= 100
2 = 50

So the arithmetic sequences are
25, 25, 25, 25 , . . . . . . . .
22, 24, 26, 28 , . . . . . . . .
19, 23, 27, 31 , . . . . . . . .
16, 22, 28, 34 , . . . . . . . .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

Ans)

i) Given , No of terms = 3 , Sum = 300 Sum

Middle term ( 2nd term) = Number of terms

= 300 = 100
3

Given, first term = 30 , d = 100 − 30 = 70
∴ Arithmetic sequence is 30, 100, 170, . . . . . . . . . .

ii) Given X1 + X2 + X3+ X4 = 300 Su m of terms
No : of pairs
Pair sum = X1 + X4 = X2 + X3 =

= 300 = 150
2

Given X1 = 30, X1 + X4 = 150

30 + X4 = 150

X4 = 150 – 30 = 120

∴ d= 120 − 30 = 120 − 30 = 90 = 30
4−1 3 3

∴ Arithmetic sequence is 30, 60, 90, 120 , . . . . . . .

iii) Given , No of terms = 5 , Sum = 300 Sum

Middle term ( 3rd term) = Number of terms

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 4

= 300 = 60
5

Given, first term = 30
Term difference
d= Position difference

∴ d= 60− 30 = 30 = 15
3− 1 2

∴ Arithmetic sequence is 30, 45, 60, 75, 90, . . . . . . . . . .

iv) Given , sum = 300,

No of terms = 6, so , No of pairs = 3
Su m of terms 300
Pair sum = No : of pairs = 3 = 100

Given X1 = 30, X1 + X6 = 100

30 + X6 = 100

70 − 30 70 − 30 X6 = 100 – 30 = 70
6−1 5 40
d= = = 5 =8

∴ Arithmetic sequence is 30, 38, 46, 54 ,62, . . . . . . .

Ans) ∴ Middle term = X3 = Su m
i) Given , S5 = 150 =
No : of terms
150
5 = 30

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 5

X1, X2, X3, X4, X5, X6, X7, X8, X9, X10

ii) Given , S10 = 550

Pair Sum = X3 + X8 5 pairs
Su m of terms
= No : of pairs

= 550 =110
5

X3 + X8 = 110

30 + X8 = 110

X8 = 110 − 30 = 80

iii) X3 = 30, X8 = 80

d= 80− 30 = 80− 30 = 50 = 10
8−3 5 5

∴ Arithmetic Sequence is 10, 20, 30, . . . . . . .

Ans )
We know Pentagon has 5 angles, and the sum of the 5 angles is
5400.

Let the smallest angle be ‘X1’ and
5 angles be X1 , X2 , X3 ,X4 ,X5

X1 + X2 + X3+ X4 + X5 = 5400

Middle term = X3 = Su m = 540 = 1080
No : of terms 5

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 6

Pair Sum = X1 + X5 = 2 × Middle term < Less than
= 2 × 1080 > Greater than
= 2160

If X1 = 360 , then X5 = 2160 − 360
= 1800 , this is not possible

If X1 < 360 , then X5 > 1800 , this is not possible

∴ X1 > 360

That is ,
the smallest angle of pentagon must be greater than 360 .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1

Online Class – X − 11 14 / 07 / 2021

1. Arithmetic Sequences – Class 9

To view class

If the number of terms of an arithmetic sequence is odd

Sum = Number of terms × Middle term

Middle Term = Sum or Middle Term = Pair Sum
2
Number of terms

If the number of terms of an arithmetic sequence is even

Sum = Number of pairs × Pair sum

Pair Sum = Sum

Number of pairs

Sum of consecutive natural numbers starting with 1
• When the number of terms are odd

Odd number of terms No of Middle Sum =

1+2+3 terms Term n × middle term

(n)

3 3 + 1=4 = 2 3×2 =6
22

1+2+3+4+5 5 5 + 1 = 6 = 3 5 × 3 = 15
1+2+3+4+5+6+7 22

7 7 + 1= 8 = 4 7 × 4 = 28
22

1+2+3+4+5+6+7+8+9 9 9 + 1 =10 =5 9 × 5 = 45
22

1 +2 + 3 + . . . . . . . . . . . . . . . .+ n n n+1 n× n+1
2 2

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

If there are odd number of terms
n+1
Middle term = 2

Sum = No of terms × Middle term
n+1
= n× 2

Sum = n(n+1)
2

• When the number of terms are even

Even number of terms No of No of Sum of a Sum =
terms pairs pair No of pairs
1+2+3+4
(n) n ×
2 Pair sum
4 4
2 =2 4 + 1 = 5 2 × 5 = 10

1+2+3+4+5+6 6 6 = 3 6 + 1 = 7 3 × 7 = 21
2

1 + 2 + 3 + 4 + 5 + 6 + 7+ 8 8 8=4 8 + 1 = 9 4 × 9 = 36
2 n + 1 n ×( n + 1)
1 + 2 + 3 + . . . . . . . . . . . . .+ n n
n 2
2

If there are even number of terms
n
Number of pairs = 2

Pair Sum = n + 1

Sum = No of pairs × Pair sum
n
= 2 × ( n + 1)

Sum = n(n+1)
2

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

Conclusion :

The sum of any number of consecutive natural numbers ,

starting with one , is half the product of the last number

and the next natural number .

That is ,

1+2+3+............+n= n ( n+ 1)
2

Activity 1

Find the following sums.

a) 1 + 2 + 3 + . . . + 100

b) 2 + 4 + 6 + . . . + 200

c) 3 + 6 + 9 + . . . + 300

d) 5 + 10 + 15 + . . . + 500

Ans)

a) 1 + 2 + 3 + . . . . . . . . + 100 = 100 (100+1)
2

= 100 x 101
2

= 50 × 101 = 5050

1 + 2 + 3 + . . . . . . . . + 100 = 5050

b) 2 + 4 + 6 + . . . . . . + 200 = 2( 1 + 2 + 3 + . . . . .+ 100)
= 2 × 5050
= 10100

c) 3 + 6 + 9 + . . . . . .+ 300 = 3( 1 + 2 + 3 + . . . . .+ 100)
= 3 × 5050
= 15150

d) 5 + 10 + 15 + . . . . . + 500 = 5( 1 + 2 + 3 + . . . . .+ 100)
= 5 × 5050
= 25250

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 4

Activity 2

Consider the sequence 6 , 11 ,16 , . . . . . . . . .

a) Write the algebraic form of the sequence.

b) Find the sum of first 100 terms of this sequence.

Ans)

a) d = 11 − 6 = 5

Here each term of the sequence is one more than

the multiples of 5 . So, algebraic form is 5n + 1

b) 6 + 11 + 16 + . . . . . . . . . . . . (100 terms)

=( 5 + 1 ) + ( 10 + 1 ) + ( 15 + 1 ) + . . . . . . . . .

= ( 5 + 10 + 15 +. . . . . . . . .) + (1 + 1 + 1+ . . . . . . . . . . . . )

100 terms 100 times

= 5 ( 1 + 2 + 3 + . . . . . . + 100 ) + 1 ×100

= 5 × 5050 + 100

= 25250 + 100 = 25350

Activity 3

Consider the sequence 4 , 9 , 14 , . . . . . . . . .

a) Write the algebraic form of the sequence.

b) Find the sum of first 100 terms of this sequence.

Ans)

a) d = 9 − 4 = 5

Here each term of the sequence is one less than

the multiples of 5 . So, algebraic form is 5n − 1

b) 4 + 9 + 14 + . . . . . . . . . . . (100 terms)

=( 5 − 1 ) + ( 10 −1 ) + ( 15 −1 ) + . . . . . . . . .

=( 5 + 10 + 15 +. . . . . . . . . . .) − (1 + 1 + 1+ . . . . . . . . . . . . )

100 terms 100 times

= 5( 1 + 2 + 3 + . . . . .+ 100) − 1 ×100

= 5 × 5050 − 1 ×100

= 25250 − 100 = 25150

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 5

Activity 4

Consider the sequence 4 , 11 , 18 , . . . . . . . . .

a) Write the algebraic form of the sequence.

b) Find the sum of first 100 terms of this sequence.

Ans)

a) d = 11 − 4 = 7

Here each term of the sequence is 3 less than

the multiples of 7 . So, algebraic form is 7n − 3

b) 4 + 11 + 18 + . . . . . . . (100 terms)

=( 7 − 3 ) + ( 14 − 3 ) + ( 21 − 3 ) + . . . . . . . . .

= (7 + 14 + 21 +. . . . . . . . . . .) − ( 3 + 3 + 3 + . . . . . . . . . . .)

100 terms 100 times

= 7( 1 + 2 + 3 + . . . . .+ 100) − 3 ×100

= 7 × 5050 − 3 ×100

= 35350 − 300 = 35050

Activity 5 :

The algebraic form of an arithmetic sequence is 10n − 4

Find the sum of first 20 terms of this sequence.

Ans)

Xn = 10n − 4
d = 10

1st term , X1 = 10 × 1 − 4 = 10 − 4 = 6

Sequence is 6, 16, 26 , . . . . . . . . . . .

Sum = 6 + 16 + 26 + . . . . . . . . . . . . .(20terms)

= (10 − 4 ) + (20 − 4 ) + ( 30 − 4 ) + . . . . . . . . .

= ( 10 + 20 + 30 +. . . . . . . . . .) − ( 4 + 4 + 4 + . . . . . . . . . . . . )

20 terms 20 times

= 10( 1 + 2 + 3 + . . . . .+ 20) − 4 × 20

= 10 × 20 x 21 − 80 = 2100 − 80 = 2020
2

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 6

or

1st term , X1 = 10 × 1 − 4 = 10 − 4 = 6
20th term , X20 = 10 × 20 − 4 = 200 − 4 = 196

20 terms of sequence are 6, 16, .. . . . . . . . 196

Sum = No of pairs × Pair sum

= 20 × (196 + 6)
2

=10 × (196 + 6) = 10 × 202 = 2020

Activity 6 :

The sum of first ‘n’ terms of an arithmetic sequence

Algebraic form of arithmetic sequence is xn = an + b
1st term = x1 = a × 1 + b = a + b
2nd term = x2 = a × 2 + b = 2 a + b
3rd term = x3 = a × 3 + b = 3 a + b
4th term = x4 = a × 4 + b = 4 a + b
5th term = x5 = a × 5 + b = 5 a + b
⸽

nth term = xn = an + b
Sum of first n terms = (a×1+b)+(a×2+b)+(a×3+b) + . . . . .+(a×n+b)

= a(1 + 2 +3 + . . . + n) + (b + b + b + . . . . .+ b)
n times
=a× n(n+1) + b×n
2

= a n(n+1) + bn
2

For the arithmetic sequence

xn = an + b

the sum of the first ‘n’ terms is

x1 + x 2 + x 3 + . . . + x n = a n (n+1) + bn
2

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 7

Activity 7 :

Calculate the difference between the sums of the first 20 terms

of the arithmetic sequences 2 , 9 , 16 , . . . . and 5 , 12 , 19 , . . . .

Ans) 2 , 9 , 16 , . . . . . . d= 9−2 =7

5 , 12 , 19 , . . . . . . d = 12 − 5 = 7

Here both the arithmetic sequences have same common

difference 7.

Difference between 5 , 12 , 19 , . . . . . . . .
corresponding terms 3 3 3 . . . . . . . . . ( 20 times )
2 , 9 , 16 , . . . . . . . . . .

∴ Difference between the sums of the first 20 terms = 20 × 3
= 60

Activity 8 :
What is the difference between the sum of the first 10 terms and
the next 10 terms of the arithmetic sequence 7 , 11 , 15 , . . . . . .
Ans)
Sequence is 7 , 11 , 15 , . . . . . . . . .
d = 11 − 7 = 4

First 10 terms X1 , X2 , X3 , . . . . . X10
Next 10 terms X11 , X12 , X13 , . . . . . X20

Difference between 10d , 10d , 10d ,. . . . . . . . 10d
corresponding terms

∴ Difference between sums = 10d + 10d+ 10d + . . . . . .(10times)
=10 ×10d
= 10 × 10 × 4
= 400

Activity 9 :
Common difference of an arithmetic sequence is 6 and the sum
of the first 20 terms is 1300 . Write down the sequence .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 8

Ans) Sum of first 20 natural numbers is
20 x 21
1 + 2 + 3 +. . . . . . . . .+ 20 = 2 = 210

Any arithmetic sequence with common difference 6 is

obtained by multiplying the sequence of natural numbers

by 6 and adding or subtracting a fixed number.

6 ×( 1 + 2 + 3 +. . . . . . . . . + 20 ) = 6 + 12 + 18 + . . . . . . .

6 + 12 + 18 + . . . . . . . = 6 × 210 = 1260

Sum of the first 20 terms of the arithmetic sequence given

in the question is 1300 and this is more than 1260.

Difference in sum = 1300 − 1260

= 40

Divide this 40 among 20 terms of the required sequence,
40
ie, 20 =2 ,

Add this 2 to each term of the sequence 6, 12, 18, . . . . . . . .
∴ Required sequence is 8, 14, 20, . . . . . . .

Assignment
T.B page 36

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1

Answers of Maths Note 09

Assignment (TB.Page 35)

Ans)

i) 11, 22, 33 , . . . . . . .

f =11 , d = 22 − 11=11

Xn =dn + f − d

=11n + 11 − 11

=11n

So, a = 11 , b = 0

Sum = a n( n+1) +bn
2

= 11 25( 25+ 1) + 0 × 25
2

= 11 x 25 x 26
2

= 11 × 25 × 13 = 3575

ii) 12, 23, 34, . . . . . . .

f =12 , d = 23 − 12 =11

Xn =dn + f − d

=11n + 12 − 11

=11n + 1

So, a = 11 , b = 1

Sum = a n( n+1) +bn
2

= 11 25( 25+ 1) + 1 × 25
2

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

= 11 x 25 x 26 + 25
2

= 11 × 25 × 13 + 25

= 3575 + 25 = 3600

iii) 21, 32, 43 , . . . . . . .

f =21 , d = 32 − 21=11

Xn =dn + f − d

=11n + 21 − 11

=11n + 10

So, a = 11 , b = 10

Sum = a n( n+1) +bn
2

= 11 25( 25+ 1) + 10 × 25
2

= 11 x 25 x 26 + 250
2

= 11 × 25 × 13 + 250

= 3575 + 250 = 3825

iv) 19, 28, 37 , . . . . . . .

f =19 , d = 28 − 19 = 9

Xn = dn + f − d

= 9n + 19 − 9

= 9n + 10

So, a = 9 , b = 10

Sum = a n( n+1) +bn
2

=9 25( 25+ 1) + 10 × 25
2

= 9 x 25 x 26 + 250
2

= 9 × 25 × 13 + 250

= 2925 + 250 = 3175

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

v) 1, 6, 11 , . . . . . . .

f =1 , d = 6 − 1 = 5

Xn = dn + f − d
= 5n + 1 − 5

= 5n − 4

So, a = 5 , b = −4

Sum = a n( n+1) +bn
2

=5 25( 25+ 1) + −4 × 25
2

= 5 x 25 x 26 − 100
2

= 5 × 25 × 13 − 100

= 1625 − 100 = 1525

Ans)
Sequence is 7 , 10 , 14 , . . . . . . . . .
d = 10 − 7 = 3

First 20 terms X1 , X2 , X3 , . . . . . X20
Next 20 terms X21 , X22 , X23 , . . . . . X40

Difference between 20d , 20d , 20d ,. . . . . . . . 20d
corresponding terms

∴ Difference between sums = 20d + 20d+ 20d + . . . . . .(20times)
= 20 × 20d
= 20 × 20 × 4
= 1600

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 4

Ans)

6 , 10 , 14 , . . . . . . d = 10 − 6 = 4

15 , 19 , 23 , . . . . . . d = 19 − 15 = 4

Here both the arithmetic sequences have same common

difference 4.

Difference between 15 ,19 , 23 , . . . . . .
corresponding terms 9 9 9 . . . . . . . . . ( 20 times )
6 , 10 , 14 , . . . . . .

∴ Difference between the sums of the first 20 terms = 20 × 9
= 180

Ans) d = 9

The first 3 digit number which is a multiple of 9 is 108

So, First term=108

The last 3 digit number which is a multiple of 9 is 999

So, Last term=999

Term difference = 999 −108 = 891 = 99 × 9

Last term = first term + 99 x common difference

= f + 99d

999 = f + 99d which is the 100th term
∴ 100th term = 999 , No: of terms =100

Sum of all multiples of 3 digit numbers=
n
S100 = 2 (first term + last term)

= 100 ( 108 + 999 )
2

= 50 × 1107 = 55350

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1

Online Class – X − 12 15 / 07 / 2021

1. Arithmetic Sequences – Class 10

To view class

The sum of first ‘n’ natural numbers

1+2+3+............+n= n(n+1)
2

The sum of first ‘n ‘ even numbers

First ‘ n’ even numbers are 2, 4, 6, . . . . . . . 2n

Sum = 2 + 4 + 6 + . . . . . . . . .+ 2n

= 2 ( 1 + 2 + 3 + . . . . . .+ n )

=2× n(n+1)
2

= n ( n + 1)

The sum of first ‘ n ’ even numbers = n (n + 1)
2 + 4 + 6 +. . . . . . . . . .+ 2n = n (n + 1)

• The sum of first 10 even numbers = 10 × 11 = 110
• The sum of first 15 even numbers = 15 × 16 = 240
• The sum of first 20 even numbers = 20 × 21 = 420
• The sum of first 50 even numbers = 50 × 51 = 2550
• The sum of first 100 even numbers = 100 × 101 = 10100

The sum of first ‘n ‘ odd numbers

First ‘n’ odd numbers are 1, 3, 5, . . . . . . . . 2n − 1

Sum = 1 + 3 + 5 + . . . . . . . + 2 n−1

= (2 − 1) + (4 − 1) + (6 − 1) + . . . . . + (2 n−1)

= 2 + 4 + 6 + . . . . . . + 2n − 1 − 1 . . . . . . . . . − 1

= (2 + 4 + 6 + . . . . . + 2n) − (1 + 1 +. . . . . . .+ 1)

n times

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

= 2(1 + 2 + 3 + . . . . + n) −1×n

=2× n(n+1) −n
2

= n2 + n − n

= n2

The Sum of first ‘ n ’ odd numbers = n2
1 + 3 + 5 + . . . . . . . . .+ 2n − 1 = n2

• The sum of first 10 odd numbers = 102 = 100
• The sum of first 15 odd numbers = 152 = 225
• The sum of first 20 odd numbers = 202 = 400
• The sum of first 50 odd numbers = 502 = 2500
• The sum of first 100 odd numbers = 1002 = 10000

• The sum of first ‘ n ’ multiples of 3

First ‘ n ’ multiples of 3 are 3, 6, 9, . . . . . . . . 3n

Sum = 3 + 6 + 9 + . . . . . . . . . + 3n

= 3 ( 1 + 2 + 3 + . . . . . .+ n )

=3× n(n+1)
2
3 3
= 2 n2 + 2 n

• The sum of first ‘ n ’ multiples of 4

First ‘ n ’ multiples of 4 are 4, 8, 12, . . . . . . . . 4n

Sum = 4 + 8 + 12 + . . . . . . . . . + 4n

= 4 ( 1 + 2 + 3 + . . . . . .+ n )

=4× n(n+1)
2

= 2n ( n + 1 )

= 2 n2 + 2 n

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

• The sum of ‘n’ consecutive terms of the arithmetic sequence

obtained by adding 1 to the multiples of 4 .

Sequence is 5 , 9 , 13 , . . . . . . . . . (4n + 1)

Sum = 5 + 9 + 13 + . . . . . . . . . . + (4n + 1)

= (4×1+1) + (4×2+1) + (4×3+1) + . . . + (4×n+1)

= 4(1 + 2 + 3 + . . . + n ) + (1 + 1+ 1+ . . . . . . + 1)

=4× n(n+1) +1×n n times
2

= 2n ( n + 1 ) + n

= 2 n2 + 2 n + n

= 2 n2 + 3 n

• The sum of first ‘n’ terms of this sequence .

The algebraic form of an arithmetic sequence is an + b .
xn = an + b = dn +(f−d)
xn = an + b
(a=d & b=f−d)
Sum = x1 + x2 + x3 + . . . . . . . . . + xn

= (a×1+b) + (a×2+b) + (a×3+b) + . . . . . .+ (a×n+b)

= a (1 + 2 + 3 + . . . . . . . + n) + (b + b + b + . . . . . . .+ b)

=a× n(n+1) +b×n n times
2
a a
= 2 n2 + 2 n+bn

= a n2 + a +b n
2 2

In this a & a + b are constants associated with the
2 2

sequence. Thus the sum is the sum of products of ‘ n2’’ and

‘n’ with definite numbers.

In other words, a a
2 2
Sum = pn2 + q n , where p = &q= +b

p= a = d (half of the common difference)
2 2
d d
p+q= 2 + 2 +f − d =d+f−d =f (the first term)

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 4

The algebraic form of the

sum of an arithmetic sequence is

p n2 + q n
d
where p = 2 and p + q = f

Q1) The algebraic form of the sum of an arithmetic sequence
is 3n2 + 4n. Find the algebraic form of the sequence ?

Ans) Given,
Sum of the first n terms = 3 n2 + 4n

Here, p = 3 , q = 4

p= d , ∴ d=p×2=3×2=6
2

p+q=f , ∴ f=3+4=7

Algebraic form of the sequence is xn = dn + f − d
=6n+7−6
=6n+1

OR
Sum of the first n terms = 3 n2 + 4n
First term = 3 × 12 + 4 × 1 = 3 × 1 + 4 = 3 + 4 = 7
Sum of the first two terms = 3 × 22 + 4 × 2

= 3 × 4 + 8 = 12 + 8 = 20
First term = 7 ,
First term + Second term = 20

Second term = 20 − First term
= 20 − 7 = 13

∴ Sequence is 7, 13, 19, . . . . . . . . . .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 5

Q2) 1

23

456

7 8 9 10

.................................

.................................

a) Write the next two lines of the pattern above .

b) How many numbers are there in the 10th line ?

c) Write the last term of the 9th line .

d) Write the First number in the 10th line .

e) Write the Last number in the 10th line .

f) Find the sum of the numbers in the 10th line .

Ans)

Line Pattern No: of terms

11 1

2 2 3 1+2 2

1+2+3 3

3 456

4 7 8 9 10 1+2+3 +4 4

...................

.........................

a) 11 12 13 14 15 (d = 1)

16 17 18 19 20 21

b) Total numbers in the 10th line = 10

c) Last number in the 9th line = 1 + 2 + 3 +. . . . . . . + 9
9 x 10
= 2 = 45

d) First number in the 10th line = 45 + 1 = 46 ( d is 1)

e) Last number in the 10th line = 1 + 2 + 3 +. . . . . . . . + 10
10 x 11
= 2 = 55

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 6

f ) Sum of the numbers in the 10th line = No of pairs × pair sum
10
= 2 ( 46 + 55 )

= 5 × 101

= 505

Q3) 3 (d = 4)
7 11
15 19 23
27 31 35 39
.................................
.................................

a) Write the next two lines of the pattern above .
b) How many numbers are there in the 10th line ?
c) Write the last term of the 9th line .
d) Write the First number in the 10th line .
e) Write the Last number in the 10th line .
f) Find the sum of the numbers in the 10th line .
Ans)
a) 35 38 41 44 47

50 53 56 59 62 65

b) Total numbers in the 10th line = 10

c) The numbers of the given pattern are in the order,
3, 7, 11, 15. . . . . . . which is an arithmetic sequence obtained
multiplying the natural numbers by 4 and subtracting 1 .

∴ Last number in the 9th line = 4 × 9 x 10 – 1 Last number in the
2 9th line of the
pattern of natural
= 180 − 1 numbers is ,

= 179 9 x 10
2

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 7

d) First number in the 10th line = 179 + 4 = 183 ( d is 4)

e) Last number in the 10th line = 4 × 10 x 11 –1 Last number in the
2 10th line of the
pattern of natural
= 220 − 1 numbers is ,

= 219 10 x 11
2

f) Sum of the numbers in the 10th line = No of pairs × pair sum
10
= 2 (183 + 219 )

= 5 × 402

= 2010

Assignment
T.B page 36

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 8

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1

Answers of Maths Note 10

Assignment (TB.Page 35 & 36)

Ans)

Algebraic form of sum = p n2 + q n
d
where p = 2 and p + q = f

i) Sum of the first n terms = n2 + 2n
Here, p = 1 , q = 2

p= d , ∴ d=p×2=1×2=2
2

p+q=f , ∴ f=1+2=3

Algebraic form of the sequence is xn = dn +( f − d)
= 2 n +( 3 − 2)

=2n+1

Sum = n2 + 2n, OR

First term ( f ) =12 + 2 × 1 = 1+ 2 = 3

Sum of first two terms = 22 + 2 × 2 = 4 + 4 = 8

Second term = 8 – 3 = 5

The arithmetic sequence is 3, 5,. . . . . . . (d = 5 – 3 = 2)

nth term = dn + (f – d) = 2n + (3 – 2)

= 2n + 1

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

ii) Sum of the first n terms = 2n2 + n
Here, p = 2 , q = 1

p= d , ∴ d=p×2=2×2=4
2

p+q=f , ∴ f=2+1=3

Algebraic form of the sequence is xn = dn + (f − d)
= 4 n +( 3 − 4)

=4n−1

iii) Sum of the first n terms = n2 − 2n
Here, p = 1 , q = − 2

p= d , ∴ d=p×2=1×2=2
2

p + q = f , ∴ f = 1 +− 2 = − 1

Algebraic form of the sequence is xn = d n + (f − d)
= 2 n + ( − 1 − 2)
= 2 n +( − 3) = 2n − 3

iv) Sum of the first ‘ n ’ terms = 2n2 − n
Here, p = 2 , q = − 1

p= d , ∴ d=p×2=2×2=4
2

p + q = f , ∴ f = 2 +− 1 = 1

Algebraic form of the sequence is xn = d n +( f − d)

= 4 n +( 1 − 4)
= 4 n +( − 3) = 4n − 3

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

iv) Sum of the first ‘ n ’ terms = n2 − n
Here, p = 1 , q = − 1

p= d , ∴ d=1×2=2
2

p + q = f , ∴ f = 1 +− 1 = 0

Algebraic form of the sequence is xn = d n +( f − d)
= 2 n +( 0 − 2)
= 2 n +( − 2) = 2n − 2

Ans)

i) Sum = 51 + 52 + 53 + . . . . . . . . + 70

= (50 + 1) + (50 + 2) + (50 + 3) +. . . . . . + (50 + 20)

= 50 + 50 + 50 +. . . . . (20 times) + (1 + 2 + 3 +. . . . . . .+ 20)
20 x 21
= 50 × 20 + 2

= 1000 + 210

= 1210

ii) Sum = 1½ + 2 ½ +3 ½ +. . . . .+12 ½

= (1+ 2+ 3+. . . . .+12) + ( ½ + ½ +. . . . . . + ½ (12 times))
12 x 13
= 2 + ½ × 12

= 6 × 13 + 6

= 78 + 6 = 84

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 4

iii)

Sum = ½ + 1 + 1½ + 2 + 2 ½ +. . . . . +12 + 12 ½

= ½ + 1 + (1 + ½ ) + 2 + (2 + ½) +. . . . . +12 + (12+ ½)

= [½ + ½ +. . . .+ ½ (13times)] + (1+ 2+......+12) + (1+ 2 +......+12)
1 12 x 13 12 x 13
= 2 × 13 + 2 + 2

= 13 + 6 × 13 + 6 × 13
2

= 6.5 + 78 + 78

= 162.5

Ans)

16, 24, 32,. . . . . . . .,

f =16 , d = 24 − 16 = 8

Xn = dn + f − d
= 8n + 16 − 8

= 8n + 8 Xn = an + b
(a+ b)2 = a2 + 2ab + b2
So, a = 8 , b = 8

Sum = a n( n+1) +bn
2

=8 n( n+1) + 8n
2

= 4 n (n+1) + 8n
= 4n2 + 4n + 8n
= 4n2 + 12n
Sum after adding 9 = 4n2 + 12n + 9

= ( 2n + 3)2

Hence it is a perfect square .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi