NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4

NCERT Solutions Class 10 Maths 
Chapter – 5 (Arithmetic Progressions) 

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 5 Arithmetic Progressions Exercise 5.4 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 5 Arithmetic Progressions

• NCERT Class 10 Maths Solution Ex – 5.1
• NCERT Class 10 Maths Solution Ex – 5.2
• NCERT Class 10 Maths Solution Ex – 5.3

Exercise – 5.4

1. Which term of the AP: 121, 117, 113, . . ., is its first negative term?
[Hint: Find n for a_n < 0]

Solution –  Given :
the AP series is 121, 117, 113, . . .,
a = 121
d = 117 – 121= -4
a_n = a + (n −1)d
a_n = 121 + (n−1)(-4)
a_n = 121 – 4n + 4
a_n = 125 – 4n
To find the first negative term of the series, a_n < 0
125 – 4n < 0
125 < 4n
n > 125/4
n > 31.25
Therefore, the first negative term of the series is the 32^nd 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 the first sixteen terms of the AP.

Solution – n^th term formula, a_n = a + (n − 1)d
Third term, a₃ = a + (3 – 1)d
a₃ = a + 2d
Seventh term, a₇ = a + (7 – 1)d
a₇ = a + 6d
According to given conditions,
a₃ + a₇ = 6         ——————– (i)
a₃ × a₇ = 8        ——————– (ii)
Substituting the values in eqn. (i), we get,
a + 2d + a + 6d = 6
2a + 8d = 6
a + 4d = 3
a = 3 – 4d        ——————– (iii)
Now substituting the values in eqn. (ii), we get,
(a + 2d) × (a + 6d) = 8      ——————– (iv)
Putting the value of a from eqn. (iii) in eqn. (iv), we get,
(3 – 4d + 2d) × (3 – 4d + 6d) = 8
(3 – 2d) × (3 + 2d) = 8
3² – (2d)² = 8         (using the identity, (a + b)(a – b) = a² – b²)
9 – 4d² = 8
4d² = 9 – 8 = 1
d = √(1/4)
d = ±1/2
d = 1/2 or -1/2
So now, if
d = 1/2,
a = 3 – 4d
a = 3 – 4(1/2)
a = 3 – 2
a = 1
and if d = -1/2,
a = 3 – 4d
a = 3 – 4(-1/2)
a = 3 + 2
a = 5
Sum of n^th term of AP is: 
S_n =
a = 1
d = 1/2
Then, the sum of first 16 terms are;
S₁₆ = 16/2 [2 + (16-1)1/2]
S₁₆ = 8(2 + 15/2) 
S₁₆ = 76
And when
a = 5
d = -1/2
Then, the sum of first 16 terms are;
S_n =
S₁₆ = 16/2 [2.5 + (16-1)(-1/2)]
S₁₆ = 8(5/2) 
S₁₆ = 20

3. A ladder has rungs 25 cm apart (see Fig. 5.7). 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 m apart, what is the length of the wood required for the rungs?
[Hint: Number of rungs = -250/25 ].

Solution –  Given:
Distance between the rungs of the ladder is 25 cm.
Distance between the top rung and bottom rung of the ladder will be (in cm) = 2 ½ × 100 = 250 cm
Hence, the total number of rungs = 250/25 + 1 = 11
As we can observe here, that, the ladder has rungs in decreasing order from top to bottom. Thus, the rungs are decreasing in an order of AP.
So, According to given condition 
a = 45
l = 25
Number of terms, n = 11
And the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed.
Now, as we know, sum of n^th terms is equal to, 
S_n =
S_n= 11/2(45 + 25)
S_n = 11/2(70)
S_n = 385 cm
Hence, the length of the wood required for the rungs is 385cm.

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 :S_{x – 1} = S₄₉ – S_x]

Solution –  Given :
Row houses are numbers from 1, 2, 3, 4, 5 ……. 49.
Thus, we can see the houses numbered in a row are in the form of AP.
a = 1
d = 1
Let us say the number of the houses can be represented as
Sum of n^th term of AP.
S_n =
Sum of number of houses beyond x house = S_{x – 1}
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(x – 1)/2            ——————— (i)
By the given condition, we can write
S₄₉ – S_x = {49/2[2(1) + (49 – 1)1]} – {x/2[2(1) + (x – 1)1]}
S₄₉ – S_x = 25(49) – x(x + 1)/2   ——————— (ii)
As per the given condition, eq.(i) and eq(ii) are equal to each other.
Therefore,

As we know, the number of houses cannot be a negative number. Hence, the value of x is 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 Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint: Volume of concrete required to build the first step = ¼ ×1/2 ×50 m³.]

Solution –  As we can see from the given figure, 
1^st step is ¼ m high, ½ m wide and 50 m long
2^nd step is (¼+¼ = ½ m) high, ½ m wide and 50 m long and, 
3^rd step is (3×¼ = 3/4 m) high, ½ m wide and 50 m long. 
Hence, we can conclude that the height of step increases by ¼ m each time when width and length is ½ m and 50 m respectively. 
So, the height of steps forms a series AP in such a way that;
¼, ½ , ¾, 1, 5/4, ……..
Volume of steps = Volume of Cuboids = Length × Breadth × Height
Now,
Volume of concrete required to build the first step = ¼ ×½ ×50 = 25/4
Volume of concrete required to build the second step =½ ×½×50 = 25/2
Volume of concrete required to build the second step = ¾ × ½ ×50 = 75/2
Volume of steps = ½ × 50 × (¼ + 2/4 + 3/4 + 4/4 + 5/4 + …….)              ——————- (i)
Now, we can see the height of concrete required to build the steps, are in AP series;
Thus, applying the AP series concept to the height,
a = 1/4
d = 2/4-1/4 = 1/4
n = 15
As, the sum of n terms is
S_n =

S₁₅ = 15/2(2×(1/4 ) + (15 – 1)1/4)
S₁₅ = 15/2 (4)
S_n = 30
Hence, up solving eqn. (1), we get
Volume of steps = ½ × 50 × S_n
= ½ × 50 × 30
= 750 m³
Hence, the total volume of concrete required to build the terrace is 750 m³.

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