# SCERT Kerala Maths Textbook solutions/Class 10/ Arithmetic Sequences – Chapter 1

SCERT Kerala Maths Textbook solutions/Arithmetic sequence is about the solutions of all the questions given in the textbook. Here you can find out textbook solutions for Class 10 Mathematics.

SCERT Kerala Maths Class 10 Text book solutions/ Arithmetic Sequences/ Chapter 1/ Part 1
Topic: Algebra of Sequences

1. Write the algebraic expression for each of the sequences below:
i) Sequence of odd numbers
ii) Sequence of natural numbers which leave remainder 1 on division by 3.
iii) The sequence of natural numbers ending in 1.
iv) The sequence of natural numbers ending in 1 or 6.
Solution:
i) Sequence of odd numbers is 1, 3, 5, 7, 9——————
Algebraic expression for this sequence is 2n – 1, where n = 1, 2, 3, 4—————-
ii) Sequence of natural numbers which leave remainder 1 on division by 3 is 1, 4, 7, 10, 13 ———-
Algebraic expression for this sequence is 3n – 2, where n = 1, 2, 3, 4—————
iii) The sequence of natural numbers ending in 1 is 1, 11, 21, 31, 41, ——————-
Algebraic expression for this sequence is 10n – 9, where n = 1, 2, 3, 4—————–
iv) The sequence of natural numbers ending in 1 or 6 is 1, 6, 11, 16, 21, ————–
Algebraic expression for this sequence is 5n – 4, where n = 1, 2, 3, 4 ————-
2. For the sequence of regular polygons starting with an equilateral triangle, write the algebraic expressions for the sequence of the sums of inner angles, the sums of the outer angles, the measures of an inner angle, and the measures of an outer angle.
Solution:
Sum of inner angles of an equilateral triangle, square, regular pentagon etc. are 180, 360, 540, 720 ——————
Algebraic expression for this sequence is 180n, where n = 1, 2, 3, 4 ————
Sum of outer angles of any polygon is 360 degree, so algebraic expression is 360 itself.
The measures of inner angles are 180/3, 360/4, 540/5 ———–
Algebraic expression for this sequence is 180n / (n + 2), where n = 1, 2, 3, 4, —————-
The measures of outer angles are 360/3, 360/4, 360/5 —————-
Algebraic expression for this sequence is 360/ (n + 2)
3. Look at the picture given in the text book (page no. 15)
The first picture is got by removing the small triangle formed by joining the midpoints of an equilateral triangle. The second picture is got by removing such a middle triangle from each of the red triangles of the first picture. The third picture shows the same thing done on the second.
i) How many red triangles are there in each picture?
ii) Taking the area of the original uncut triangle as 1, compute the area of a small triangle in each picture.
iii) What is the total area of all the red triangles in each picture?
iv) Write the algebraic expressions for these three sequences obtained by continuing this process.
Solution:
i) There are 3 red triangles in first picture, 9 in the second picture and 27 in the third picture.
ii) Area of the original uncut triangle = 1
Area of a small triangle in the first picture is ¼.
Area of a small triangle in the second picture = 1/16
Area of a small triangle in the third picture = 1/64
iii) Total area of all the red triangles in the first picture = 3 x ¼ = ¾
Total area of all the red triangles in the second picture = 9 x 1/16 = 9/16
Total area of all the red triangles in the third picture = 27 x 1/64 = 27/64.
iv) First sequence is 3, 9, 27 ————–
Algebraic expression for this sequence is 3 raised to the power n, where n = 1, 2, 3 ——-
Second sequence is ¼, 1/16, 1/64 ———————-
Algebraic expression for this sequence is (1/4) raised to the power n, where n = 1, 2, 3——
Third sequence is ¾, 9/16, 27/64 ——————–
Algebraic expression for this sequence is (3/4) raised to the power n, where n= 1, 2, 3, ——–