# Arithmetic Sequences /SCERT Kerala Mathematics

Arithmetic Sequences – Chapter 1

## SCERT Kerala Syllabus Mathematics Worksheets

- Consider the arithmetic sequence 3, 8, 13, 18—————

a) Find its common difference?

b) Find its 25th term?

Answer:

a) Given a = 3

Common difference = 8 – 3 = 5

b) 25th term = a + (n – 1) d = 3 + 24 x 5 = 123 - 6th term of an arithmetic sequence is 42 and its 10th term is 70.

a) Find the common difference?

b) What is its first term?

Answer:

a) Common difference = (70 – 42)/ (10 – 6) = 28/4 = 7

b) Given 6th term = 42

6th term = a + 5d = 42

a + 5 x 7 = 42

a + 35 = 42

a = 42 – 35 = 7 - The algebraic form of an arithmetic sequence is 2n + 3.

a) Find its common difference?

b) What is its first term?

Answer:

a) When n = 1, 2 x 1 + 3 = 5

When n = 2, 2 x 2 + 3 = 7

When n = 3, 2 x 3 + 3 = 9 ———

Arithmetic sequence is 5, 7, 9, ———-

a) Common difference = 7 – 5 = 2

b) First term = 5 - Find the following sums:

a) 1 + 2 + 3 + ——- + 20

b) 2 + 4 + 6 + ——– + 20

c) 1 + 3 + 5 + ——– + 20

Answer:

a) n (n + 1) /2 = (20 x 21)/2 = 210

b) Given a = 2, d = 2

Sum = n/2[2a + (n-1) d] = 20/2 [2 x 2 + (20 – 1)2]

= 10 [4 + 19 x 2]

= 10 [4 + 38] = 10 x 42 = 420 - Consider an arithmetic sequence of algebraic form 2n + 5

a) Find the common difference?

b) Find its first term?

Answer:

a) When n = 1, 2n + 5 = 2 x 1 + 5 = 7

When n = 2, 2 x 2 + 5 = 9

When n = 3, 2 x 3 + 5 = 11 ———

So the arithmetic sequence is 7, 9, 11, ———–

a) Common difference = 9 – 7 = 2

b) First term = 7 - Find the sum of

a) 1 + 2 + 3 + ——- + 200

b) 4 + 8 + 12 + ——— + 800

Answer:

a) n (n + 1)/2 = (200 x 201)/2 = 20100

b) 4 + 8 + 12 + —— + 800 = 4( 1 + 2 + 3 + —-+ 200)

= 4 x 20100 = 80400 - Write the algebraic expression for the sequence of natural numbers ending in 1.

Answer:

Required sequence is 1, 11, 21, 31, ————

1 = 10 x 1 – 9

11 = 10 x 2 – 9

21 = 10 x 3 – 9

Algebraic expression is 10n – 9, for n = 1, 2, 3——————- - The first term of an arithmetic sequence is 1 and the sum of the first four terms is 100. Find the first four terms.

Answer:

First term = 1

Let the first four terms be f, f+d, f+2d, f+3d.

Given f + f + d + f + 2d + f + 3d = 100

4f +6d = 100 or 2f + 3d = 50

2 + 3d = 50

3d = 48

d = 48/3 = 16

The arithmetic sequence is 1, 17, 33, 49, ———— - Write four arithmetic sequences with 200 as the sum of the first four terms.

Answer:

Let the first four terms be x – 3d, x – d, x + d, x + 3d

Given x – 3d + x – d + x + d + x + 3d = 200

4x = 200

x = 200/4 = 50

i) If d = 1, x – 3d = 50 – 3 = 47

x – d = 50 – 1 = 49

x + d = 50 + 1 = 51

x + 3d = 50 + 3 = 53

The arithmetic sequence is 47, 49, 51, 53, ——–

ii) If d = 2, x – 3d = 50 – 6 = 44

x – d = 50 – 2 = 48

x + d = 50 + 2 = 52

x + 3d = 50 + 6 = 56

The arithmetic sequence is 44, 48, 52, 56, ————

iii) If d = 3, x – 3d = 50 – 9 = 41

x – d = 50 – 3 = 47

x + d = 50 + 3 = 53

x + 3d = 50 + 9 = 59

The arithmetic sequence is 41, 47, 53, 59, ——–

iv) If d = 4, x – 3d = 50 – 12 = 38

x – d = 50 – 4 = 46

x + d = 50 + 4 = 54

x + 3d = 50 + 12 = 62

The arithmetic sequence is 38, 46, 54, 62, ———– - Find the sum of all three digit numbers which are multiples of 9.

Answer:

Required sequence is 108, 117, 126, ———, 999

Number of three digit numbers = (999 – 108)/9 + 1 = 100

Sum = n/2(first term + last term)

= 100/2(108 + 999)

= 50 x 1107 = 55350