# SCERT Kerala Maths Class 10/Arithmetic Sequences

SCERT Kerala Maths Class 10/ Arithmetic sequence is about the textbook solutions of Class 10 Mathematics of Chapter 1 Arithmetic Sequences. Here you can find out practice problems of Class 10 Mathematics .

SCERT Kerala Maths Class 10 Textbook Solutions/ Arithmetic Sequences
Kerala State Board SSLC Arithmetic Sequences – Chapter 1
Sums and Terms:

1. Write three arithmetic sequences with 30 as the sum of the first five terms.
Solution:
Let the first 5 terms be x-2, x-1, x, x+1, x+2
Now x – 2 + x – 1 + x + x + 1 + x + 2 = 30
5x = 30
x = 30/5 = 6
The arithmetic sequence is 4, 5, 6, 7, 8.
Let the first 5 terms be x, x+2, x+4, x+6, x+8.
Now x + x + 2 + x + 4 + x + 6 + x + 8 = 30
5x + 20 = 30
5x = 30 – 20 = 10
x = 10/5 = 2
The arithmetic sequence is 2, 4, 6, 8, 10
Let the first 5 terms be x, x+1, x+2, x+3, x+4
Now x + x + 1 + x +2 + x + 3 + x + 4 = 30
5x + 10 = 30
5x = 30 – 10 = 20
x = 20/5 = 4
The arithmetic sequence is 4, 5, 6, 7, 8
2. The first term of an arithmetic sequence is 1 and the sum of the first four terms is 100. Find the first four terms.
Solution:
1st term = 1
Let the first 4 terms be x, x +d, x+2d, x+3d
Given x + x + d + x + 2d + x + 3d = 100
4x + 6d = 100 or 2x + 3d = 50
But first term is 1(given), so 2 + 3d = 50
3d = 50 – 2 = 48
d = 48/3 = 16
Arithmetic sequence is 1, 17, 33, 49, ———————–
3. Prove that for any four consecutive terms of an arithmetic sequence, the sum of the two terms on the two ends and the sum of the two terms in the middle are the same.
Solution:
Let 4 consecutive terms of an arithmetic sequence is x, x+ d, x+ 2d, x + 3d
Sum of the 2 terms on the two ends = x + x + 3d = 2x + 3d
Sum of the two terms in the middle = x + d + x + 2d = 2x + 3d
So both are same.
4. Write four arithmetic sequences with 100 as the sum of the first four terms.
Solution:
Let the first 4 terms be x – 3d, x –d, x + d, x + 3d
Given x – 3d + x – d + x + d + x + 3d = 100
4x = 100
x = 100/4 = 25
Put d = 1, 2, 3, 4————
When d = 1, x – 3d = 25 – 3 = 22
x – d = 25 – 1 = 24
x + d = 25 + 1 = 26
x + 3d = 25 + 3 = 28
So the arithmetic sequence is 22, 24, 26, 28 —————-
When d = 2, x – 3d = 25 – 3 x 2 = 25 – 6 = 19
x – d = 25 – 2 = 23
x + d= 25 + 2 = 27
x + 3d = 25 + 3 x 2 = 25 + 6 = 31
So the arithmetic sequence is 19, 23, 27, 31 ————–
When d = 3, x – 3d = 25 – 3 x 3 = 25 – 9 = 16
x – d = 25 – 3 = 22
x + d = 25 + 3 = 28
x + 3d = 25 + 3 x 3 = 25 + 9 = 34
So the arithmetic sequence is 16, 22, 28, 34, ————–
When d = 4, x – 3d = 25 – 3 x 4 = 25 – 12 = 13
x – d = 25 – 4 = 21
x + d = 25 + 4 = 29
x + 3d = 25 + 3 x 4 = 25 + 12 = 37
So the arithmetic sequence is 13, 21, 29, 37, —————-
5. Write the first three terms of each of the arithmetic sequences described below:
i) First term 30; the sum of the first three terms is 300
ii) First term 30; the sum of the first four terms is 300
iii) First term 30; the sum of the first five terms is 300
iv) First term 30; the sum of the first six terms is 300
Solution:
i) We know that, the sum of any three consecutive terms of an arithmetic sequence is 3 times the middle term.
Let the middle term be x
Then given 3x = 300
x = 300/3 = 100
First term is 30 and middle term is 100
So common difference = 100 – 30 = 70
Third term = 100 + 70 = 170
Therefore, first 3 terms of an arithmetic sequence are 30, 100, 170, —————
ii) For the 4 consecutive terms of an arithmetic sequence, the sum of the first term and the last term is equal to the sum of the second term and the third term.
i.e. 1st + 4th = 2nd + 3rd = 300/2 = 150
Since 1st term = 1, 4th term = 150 – 30 = 120
Now the sequence is 30, —, —-, 120
Here term difference = 120 – 30 = 90
Position difference = 4 – 1 = 3
Common difference = 90/3 = 30
So the sequence is 30, 60, 90, 120 ——–
iii) Sum of the five consecutive terms of an arithmetic sequence is 5 times of its middle term.
Let the middle term be x
So 5x = 300
x = 300/5 = 60
Let the 5 consecutive terms be x – 2d, x – d, x + d, x +2d
1ST term = x – 2d = 30 (given)
60 – 2d = 30
30 = 2d
d = 30/2 = 15
Therefore, the sequence is 30, 45, 60, 75, ——————–
iv) For the six consecutive terms of an arithmetic sequence,
1st term + 6th term = 2nd term + 5th term = 3rd term + 4th term = 300/3 = 100
Since 1st term =30, 6th term = 100 – 30 = 70
Now term difference = 70 – 30 = 40
Position difference = 6 – 1 = 5
Common difference = 40/5 = 8
So the sequence is 30, 38, 46, 54, 62, 70, ———————-
6. The sum of the first five terms of an arithmetic sequence is 150 and the sum of the first ten terms is 550.
i) What is the third term of the sequence?
ii) What is the eighth term?
iii) What are the first three terms of the sequence?
Solution:
i) Sum of first 5 terms = 150
5 times its middle term 5x = 150
x = 150/5 = 30
So middle term = third term = 30
ii) Sum of the first 10 terms = 550
1st + 10th = 2nd + 9th = 3rd + 8th = 4th + 7th = 5th + 6th = 550/5 = 110
Since 3rd term = 30, 8th term = 110 – 30 = 80
iii) Now 3rd term = 30 and 8th term = 80
Term difference = 80 – 30 = 50
Position difference = 8 – 3 = 5
Common difference = 50/5 = 10
So the sequence is 10, 20, 30, 40, 50, —————
7. The angles of a pentagon are in arithmetic sequence. Prove that its smallest angle is greater than 36 degree.
Solution:
Let the angles of a pentagon are x, x +d, x+2d, x+3d, x+4d
Sum of all angles of a pentagon = (n – 2) 180 = (5 – 2) 180 = 3 x 180 = 540 degree.
Now x + x + d + x + 2d + x + 3d + x + 4d = 540
5x + 10d = 540
x + 2d = 108
If we take the smallest angle x as 36 , 36 + 2d = 108
2d = 108 – 36 = 72
d = 72/2 = 36
So the 5 angles are 36, 72, 108, 144, 180.
Here 180 degree cannot be an angle of a pentagon, so its smallest angle will be greater than 36 degree.