# SCERT Kerala Maths Class 10/ Arithmetic Sequences.

SCERT Kerala Maths Class 10/ Arithmetic Sequence is about the textbook solutions of Class 10 Mathematics – Arithmetic Sequences. Here you can find out practice problems from the chapter Arithmetic Sequences.

SCERT Kerala Maths Class 10 Textbook Solutions/ Arithmetic Sequences
Kerala State Board SSLC /Arithmetic Sequences – Chapter 1
Algebra of Arithmetic Sequences:

1. The 8th term of an arithmetic sequence is 12 and its 12th term is 8. What is the algebraic expression for this sequence?
Solution:
Given 8th term = 12 and 12th term = 8
To get 12th term from 8th term, we must add the common difference 12 – 8 = 4 times
Term difference = 8 – 12 = -4
Thus 4 times common difference = -4
Common difference = -4/4 = -1
The algebraic form of any arithmetic sequence is of the form an + b, where a and b are fixed numbers, a is the common difference.
So we can write 12 = -1 x 8 + b
b = 12 + 8 = 20
Or 8 = -1 x 12 +b
b = 8 + 12 = 20
The algebraic expression is xn = -1 x n + 20 = 20 – n
2. The Bird problem in Class 8 (the lesson, Equations) can be slightly changed as follows.
One bird said:
“We and we again, together with half of us and half of that, and one more is a natural number”
Write all the possible number of birds, starting from the least. For each of these, write the sum told by the bird also. Find the algebraic expression for these two sequences.
Solutions:
Let us take the number of birds as x.
Given x + x + x/2 + x/4 + 1 is a natural number.
Therefore, 2x + 3x/4 + 1 = (11x + 4)/4 is a natural number
i.e. 11x + 4 is a multiple of 4
i.e. 11x is a multiple of 4
i.e. x is a multiple of 4
So the number of birds is 4, 8, 12, 16—————–
Now (11x + 4)/4 when x = 4, 8, 12, 16 —————— are 12, 23, 34, 45, ————-
Algebraic expression is f + (n- 1) d = 12 + (n – 1) 11
= 12 + 11n – 11 = 1 + 11n
3. Prove that the arithmetic sequence with first term 1/3 and common difference 1/6 contains all natural numbers.
Solution:
Given 1st term = 1/3 and Common difference = 1/6
Also n th term = f + (n – 1) d = 1/3 + (n – 1) 1/6 = 1/3 + (1/6) n – 1/6
= 1/6 + (1/6) n = 1/6(1 + n)
If we are giving the values of n as 5, 11, 17, ——— we will get all natural numbers.
So arithmetic sequence with first term 1/3 and common difference 1/6 contains all natural numbers.
4. Prove that the arithmetic sequence with first term 1/3 and common difference 2/3 contains all odd numbers, but no even number.
Solution:
Given 1st term = 1/3 and Common difference = 2/3
Also n th term = f + (n-1) d = 1/3 + (n -1) 2/3 = 1/3 + 2n/3 – 2/3
= 2n/3 – 1/3
= 1/3(2n -1)
Here value of 2n – 1 is always an odd number, so it does not contain any even number.
5. Prove that the squares of all the terms of the arithmetic sequence 4, 7, 10, ———belong to the sequence.
Solution:
Here 1st term = 4 and common difference = 3
n th term = f + (n – 1) d = 4 + (n – 1) 3 = 4 + 3n – 3 = 3n + 1
Square of (3n + 1) for different values of n also includes this sequence
6. Prove that the arithmetic sequence 5, 8, 11 , —————- contains no perfect squares.
Solution:
Here 1st term = 5 and Common difference = 3
Thus n th term = f + (n – 1) d = 5 + (n-1) 3 = 5 + 3n – 3 = 3n +2
We know that when we divide a perfect square by 3, we will get the remainder as 1 or 0. But here if we divide 3n + 2 by 3, we get the remainder as 2. So it does not contain any perfect squares.
7. Write the whole numbers in the arithmetic sequence 11/8, 14/8, 17/8, ————————
Do they form an arithmetic sequence?
Solution:
Here 1st term = 11/8 and Common difference = 3/8
Also n th term = f + (n-1) d = 11/8 + (n – 1) 3/8
= 11/8 + 3n/8 – 3/8
= 8/8 + 3n/8
= 1 + 3n/8
If we took the value of n as multiple of 8, we get 4, 7, 10, ———–as a sequence of whole numbers with common difference 3.So it forms an arithmetic sequence.