# SCERT Kerala Class 10 Mathematics/Arithmetic Sequences.

**SCERT Kerala Class 10 Mathematics/Arithmetic Sequences is about some extra questions for practice. Here you can find out practice problems for Class 10 Mathematics.**

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

**Write the sequence obtained by subtracting 2 from the multiples of 3? Is this an arithmetic sequence? Also write the algebraic form?**

Solution:

3 x 1 – 2 = 1

3 x 2 – 2 = 4

3 x 3 – 2 = 7

3 x 4 – 2 = 10 ————–

So the sequence obtained is 1, 4, 7, 10, ——————-

Now common difference = 4 – 1 = 3, 7 – 4 = 3, 10 – 7 = 3, ——-

Since the difference between two consecutive terms is a constant, 1, 4, 7, 10 ——— is an arithmetic sequence.

Algebraic form is 3n – 2.**5th term of an AP is 43 and common difference is 4, find the first term and 12th term?**

Solution:

Given 5th term = 43, d = 4.

First term = 5th term – 4 d = 43 – 4 x 4 = 43 – 16 = 27

12th term = 5th term + 7d = 43 + 7 x 4 = 43 + 28 = 71.**The 7th term of an AP is 46 and 10th term is 61.**

a) Find the common difference?

b) Find the 15th term?

Solution:

Common difference = Term difference /Position difference = (61 – 46)/ (10 – 7) = 15/3 = 5

15th term = 7th term + 8d = 46 + 8 x 5 = 46 + 40 = 86**The 5th term of an AP is 52 and 7th term of an AP is 64. Find the 6th term?**

Solution:

Given 5th term = 52

7th term = 64

6th term = ½ (5th term + 7th term) = ½ (52 + 64) = 116/2 = 58.**Find the algebraic form of the sequence 18, 24, 30, 36, ————**

Solution:

d = 24 -18 = 6 & f = 18

xn = f + (n-1)d = dn + f – d = 6n + 18 – 6 = 6n + 12**The n th term of an arithmetic sequence is 3n – 7. Find the 5th term?**

Solution:

5th term = 3 x 5 – 7 = 15 – 7 = 8**How many terms are there in the sequence 2, 5, 8, 11, 14, ———-, 119.**

Solution:

Given dn + f – d = 119

3n + 2 -3 = 119

3n – 1 = 119

3n = 120

n = 120/3 = 40

So 40 terms are there in the sequence.**Is 200 a term of the arithmetic sequence 2, 8, 14, 20, 26, —————–?**

Solution:

Let us take xn as 200.

dn + f – d = 200

6n + 2 – 6 = 200

6n – 4 = 200

6n = 204

n = 204/6 = 34

So 34th term is 200.**Find the sum of the arithmetic sequence 5, 8, 11, 14, ———-, 152?**

Solution:

Here 1st term = 5 & last term = 152, d= 3

Now number of terms = [(last term – first term)/d] + 1

= (152 – 5)/3 + 1 = 147/3 + 1 = 49 + 1 = 50

Sum = n/2(first term + last term) = 50/2(5 + 152) = 25 x 157 = 3925**Find the sum of first 50 terms of the sequence 1, 4, 7, 10, —————**

Solution:

Given f = 1, d=3, n= 50

Sum = n/2(2f + (n-1) d) = 50/2(2 x 1+ (50-1)3) = 25(2 + 49 x 3) = 25 (2 + 147)

= 25 x 149 = 3725