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NCERT New Syllabus Mathematics
Class: 10
Exercise 5.4
Topic: Arithmetic Progressions
Introduction to Arithmetic Progressions
Arithmetic Progressions (AP) are a fundamental mathematical concept that forms the basis for understanding various real-world applications, ranging from financial planning to engineering design. In this chapter of the 10th-grade NCERT syllabus, we delve into sequences where the difference between consecutive terms remains constant. This simple yet powerful concept helps students recognize patterns and solve problems involving future predictions, geometric designs, and even natural phenomena.
An arithmetic progression represented as a sequence of numbers like a, a+d, a+2d, and so on, is defined by the first term (a) and the common difference (d). This progression opens up a range of possibilities for solving complex problems by understanding the behavior of numbers in a linear format.
In this blog, we’ll explore the key concepts of AP, including its general form, the formula for the nth term, and the sum of n terms. We’ll also solve various problems from the new NCERT syllabus, making it easier for students to grasp the concept and excel in their exams.
1) The nth term an of the AP with first term a and common difference d is given by
an = a + (n – 1) d.
2) an is also called the general term of the AP. If there are m terms in the AP, then
am represents the last term which is sometimes also denoted by l.
EXERCISE 5.4
1. Which term of the AP: 121, 117, 113, . . ., is its first negative term?
[Hint: Find n for an < 0]
Solution:
1) According to the problem, a1 = 121, a2 = 117, a3 = 113. . . so d = – 4,
2) We will have to find n using the above information.
an = a + (n – 1) d
an = 121 + (n – 1) (– 4)
an = 121 + (– 4n + 4)
an = 121 – 4n + 43) We have to find which first term is negative.
an = 125 – 4n
an < 0
125 – 4n < 0
125 < 4n
4n > 125
n > 125/4
n > 31.25
4) Therefore, the first negative term of this AP is the 32nd term.
2. The sum of the third and the seventh terms of an AP is 6 and their product
is 8. Find the sum of first sixteen terms of the AP.
Solution:
1) Let the first term be "a" and the common difference be "d".
2) Here a3 + a7 = 6 ---------- equation 1
3) Here a3 x a7 = 8 ---------- equation 2
4) Now we will find a3 and a7,
a) First we will find a3
an = a + (n – 1) da3 = a + (3 – 1) d
a3 = a + 2d ---------- equation 3
b) Now we will find a7
an = a + (n – 1) da7 = a + (7 – 1) d
a7 = a + 6d ---------- equation 4
5) From equations 1, 3, and 4, we have,
a3 + a7 = 6
a + 2d + a + 6d = 66) From equations 2, 3, 4, and 5, we have,
2a + 8d = 6
2(a + 4d) = 6
(a + 4d) = 6/2
(a + 4d) = 3
a = 3 – 4d ---------- equation 5
a3 x a7 = 8(a + 2d) x (a + 6d) = 8
(3 – 4d + 2d) x (3 – 4d + 6d) = 8
(3 – 2d) x (3 + 2d) = 8
(32 – 4d2) = 8(9 – 4d2) = 8
4d2 = 9 – 8
4d2 = 1
d2 = 1/4
d2 = 1/4
7) Put d = 1/2 and d = - 1/2 from equation 6 in equation 5,d = ± 1/2
d = 1/2 or – 1/2 ---------- equation 6
a) First we take d = 1/2, we get
a = 3 – 4da = 3 – 4(1/2)a = 3 – 2
a = 1 ---------- equation 7
b) First we take d = – 1/2, we get
a = 3 – 4da = 3 – 4(– 1/2)a = 3 – (– 2)
a = 3 + 2
a = 5 ---------- equation 8
8) We know that the sum of the first n terms of an AP is given by:
Sn = (n/2)[2a + (n – 1) d]
a) First we will find sum of first 16 terms with a = 1 and d = 1/2:Sn = (n/2)[2a + (n – 1) d]
S16 = (16/2)[2(1) + (16 – 1) (1/2)]
S16 = 8[2 + (15) (1/2)]
S16 = 8[2 + (15/2)]
S16 = 8[(4 + 15)]/2
S16 = 4(19)
S16 = 76 ---------- equation 9
b) First we will find sum of first 16 terms with a = 5 and d = (– 1/2):Sn = (n/2)[2a + (n – 1) d]
S16 = (16/2)[2(5) + (16 – 1) (– 1/2)]
S16 = 8[10 + (15) (– 1/2)]
S16 = 8[10 – (15/2)]
S16 = 8[(20 – 15)]/2
S16 = 4(5)
S16 = 20 ---------- equation 10
9) From equations 9 and 10, we have,
a) S16 = 76 when a = 1 and d = 1/2
b) S16 = 20 when a = 5 and d = – 1/2.
3. A ladder has rungs 25 cm apart. (see the following fig.). The rungs decrease
uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2½ m apart, what is the length of the wood required for the rungs?
[Hint : Number of rungs = (250/25) + 1]
Solution:
1) The distance between the rungs is 25 cm.
2) Distance between the top rung and the bottom
rung is 2½ m. i.e. 5/2 m = 2.5 m = 250 cm.
3) So total number of rungs = [(250/25) + 1] = 11.
4) The length of rungs is decreasing uniformly from bottom to top, so they are in AP.
5) Here, a1 = 45, l = 25 and n = 11, so,
Sn = (n/2)[a + l]
S11 = (11/2)[45 + 25]
S11 = (11/2)[70]
S11 = 11(35)
S11 = 385
6) Therefore, the length of the wood required for the rungs is 385 cm.
4. The houses of a row are numbered consecutively from 1 to 49. Show that
there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint : sx - 1 = S49 – Sx]
Solution:
1) Row houses are numbered 1, 2, 3, . . 49.
2) So, a1 = 1, a2 = 2, a3 = 3. . . so d = 1,
3) The sum of the number of houses preceding the xth house will be S(x-1).
Sn = (n/2)[2a + (n – 1) d]
S(x-1) = ((x – 1)/2)[2(1) + (x – 1 – 1) (1)]
S(x-1) = ((x – 1)/2)[2 + (x – 2)]
S(x-1) = ((x – 1)/2)[x]
S(x-1) = x(x – 1)/2 ---------- equation 1
Sn = (n/2)[2a + (n – 1) d]S49 = (49/2)[2(1) + (49 – 1) (1)]S49 = (49/2)[2 + 48]
S49 = (49/2)(2)[1 + 24]
S49 = (49)[25]
S49 = 1225 ---------- equation 2
5) Now we will find Sx
Sn = (n/2)[2a + (n – 1) d]Sx = (x/2)[2(1) + (x – 1) (1)]Sx = (x/2)[2 + (x – 1)]Sx = (x/2)[x + 1]Sx = x(x + 1)/2 ---------- equation 3
6) Now we will find S49 – Sx using equations 2 and 3.
S49 – Sx = 1225 – [x(x + 1)/2] ---------- equation 4
7) According to the problem, S(x-1) = S49 – Sx ---------- equation 5
8) From equations 1, 4, and 5, we have,
S(x-1) = S49 – Sx
x(x – 1)/2 = 1225 – [x(x + 1)/2]
x(x – 1)/2 = [2450 – (x(x + 1))]/2x(x – 1) = [2450 – (x(x + 1))]
x2 – x = 2450 – x2 – x
2x2 = 2450
x2 = 1225
x = ± 35
x = 35 or – 359) Therefore, the number of houses can't be negative, the value of
x will be 35.
5. A small terrace at a football ground comprises of 15 steps each of which is
50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m. (see the following fig.). Calculate the total volume of concrete required to build the terrace. [Hint: Volume of concrete required to build the first step = (1/4) x (1/2) x 50 m3]
1) According to the problem and the figure, we have,
i) The heights of the steps are given bellow:
a) The height of the first step is (1/4) mb) The height of the second step is (1/4) + (1/4) = (1/2) mc) The height of the 3rd step is (1/2) + (1/4) = (3/4) md) The height of the 4th step is (3/4) + (1/4) = (1) m
2) If we consider the height in the changing form, the width will be the same for all
the steps. i.e. the width will be 1/2 m for all the steps.
3) Here the length of the terrace is 50 m.
4) Here we can find:
The volume of the steps = volume of the cuboid
The volume of the steps = Length x Breadth x Height
5) Now we will find the volumes of the steps:
a) The volume of the first step
= (1/4) x (1/2) x (50)
= (1/8) x (50)
= (50/8)
= (25/4) ---------- equation 1
b) The volume of the second step
= (1/2) x (1/2) x (50)
= (1/4) x (50)
= (50/4) ---------- equation 2
c) The volume of the 3rd step
= (3/4) x (1/2) x (50)
= (3/4) x (25)
= (75/4) ---------- equation 3
d) The volume of the 4th step is (3/4) + (1/4) = (1) m
= (1) x (1/2) x (50)= (1) x (25)= (25) ---------- equation 4
6) From equations 1, 2, 3, and 4, volumes of the steps are in AP.
7) So, here, a = 25/4, d = 50/4 – 25/4 = 25/4, and n = 15,
8) so the sum of these 15 steps will be,
Sn = (n/2)[2a + (n – 1) d]
S15 = (15/2)[2(25/4) + (15 – 1) (25/4)]
S15 = (15/2)[2(25/4) + (14) (25/4)]
S15 = (15/2) x (2(25/4)) x [1 + (7)]
S15 = (15/2) x (2(25/4)) x [8]
S15 = [(15 x 2 x 25)/8] x [8]
S15 = (15 x 2 x 25)
S15 = (30 x 25)
S15 = (750)
9) The concrete required to build the terrace is 750 m3.
Conclusion: The Power of Arithmetic Progressions
As we wrap up our journey through Arithmetic Progressions, we've seen how sequences unfold and reveal patterns in everyday life—from predicting future events to organizing data. The concept of common differences and the sum of terms allows us to model various scenarios in both math and real-world situations. Whether you’re solving problems in exams or exploring the magic of number sequences, AP knowledge is a powerful tool. Keep practicing, and you’ll see how every sequence leads to new possibilities!
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