# SCERT Kerala Class 10 Mathematics/ Arithmetic Sequences.

**SCERT Kerala Class 10 Mathematics / Arithmetic Sequences is about the extra questions for practice for SSLC students. Here you can find out practice problems/Model questions for the Chapter Arithmetic Sequences.**

**SCERT Kerala Class 10 Mathematics /Arithmetic Sequences – Chapter 1Extra Questions for Practice/ Model QuestionsAnswer the following:**

**Consider the arithmetic sequence 11, 22, 33, ————————**

a) What is its common difference?

b) Find the algebraic form?

Solution:

a) Common difference = 22 – 11 = 11

b) Here f= 11, d= 11

Algebraic form is f + (n-1) d = 11 + (n-1)11 = 11 + 11n – 11 = 11n**Write the sequence of even numbers greater than 2. Find its algebraic form?**

Solution:

Sequence of even numbers greater than 2 is 4, 6, 8, 10, —————–

Algebraic form is f + (n -1) d = 4 + (n-1) 2 = 4 + 2n – 2 = 2n + 2 = 2(n + 1).**Is 1917 a term of the sequence 5, 9, 13, ——–?**

Solution:

Let nth term be 1917.Given f = 5, d = 4

So dn + f – d = 4n + 5 – 4 = 1917

4n + 1 = 1917

4n = 1917 – 1 = 1916

n = 1916/4 = 479

Therefore, 1917 is the 479th term.**Consider an arithmetic sequence 22/3, 28/3, 34/3, ——————**

a) Find the common difference?

b) Write the algebraic expression of the sequence?

c) Is there any counting number in this sequence? Justify your answer?

Solution:

a) Common difference = 28/3 – 22/3 = 6/3 = 2

b) Algebraic expression is f + (n – 1) d = 22/3 + (n – 1) 2

= 22/3 + 2n – 2

= 2n + 16/3

c) Here 2n is a counting number, but if we are adding 16/3 to 2n, it will become a fraction. So there won’t be any counting number in this sequence.**Find the 10th term of an arithmetic sequence if its 7th term is 18 and 18th term is 7?**

Solution:

Given 7th term = 18 and 18th term = 7

Now 18th – 17th = (18-7) d

7 – 18 = 11d

-11 = 11d

d = 11/ (-11) = -1

10th term = 7th term + 3d = 18 + 3 (-1) = 18 -3 = 15**The algebraic form of an arithmetic sequence is 5n + 2**

a) What is the first term?

b) Find its common difference?

c) Write the arithmetic sequence?

d) What is the remainder when each term is divided by 5?

Solution:

a)First term = 5 x 1 + 2 = 7

b) Common difference = 5

c) Arithmetic sequence is 7, 12, 17, 22, —————-

d) When each term is divided by 5, we will get remainder as 2.**How many terms are there in the sequence 1, 4, 7, 10, —————– 100?**

Solution:

n th term = dn + f – d = 100, d = 3, f = 1

3n + 1 – 3 = 100

3n – 2 = 100

3n = 102

n = 102/3 = 34

So there are 34 terms in the given sequence.**Compute the 25th term of the sequence 19, 28, 37, —————–**

Solution:

Here common difference = 28 – 19 = 9

25th term = 1st term + 24d = 19 + 24 x 9 = 235.**Write an arithmetic sequence with 3rd term 37 and 7th term 73?**

Solution:

Given 3rd term = 37, 7th term = 73

Now 7th term – 3rd term = (7 – 3) d

73 – 37 = 4d

36 = 4d

d = 36/4 = 9

1st term = 3rd term – 2d = 37 – 2 x 9 = 37 – 18 = 19

So the required arithmetic sequence is 19, 28, 37, ————-**Compute the sum of the first 100 terms of the arithmetic sequence 1, 4, 7, ———?**

Solution:

100th term = 1st term + 99d = 1 + 99 x 3 = 298

Now sum of first 100 terms = ½ x 100 x (1 + 298) = 14950.

Its very effective