Class 10 Maths Chapter 5 Exercise 5.4 - Arithmetic Progression NCERT Solutions
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.4 are provided here to help students prepare effectively for the CBSE Board exams. These solutions for Class 10 Maths Exercise 5.4 of Chapter 5 Arithmetic Progression are the ideal study material for clearing doubts and understanding the concepts of Arithmetic Progression.
The problems in Exercise 5.4 are particularly useful for understanding how to predict quantities in fields like finance, commerce, science, and other areas where values are arranged in a sequence. These solutions have been solved by an experienced CBSE Maths teacher, following a detailed, step-by-step approach according to CBSE standards, ensuring that every student can easily comprehend the solutions for Class 10 Chapter 5 Maths Exercise 5.4.
- All unsolved questions in Exercise 5.4 are covered.
- Solutions are provided in a detailed, step-by-step manner.
- These NCERT solutions offer insights into solving similar questions effectively.
- Exercise 5.4 includes questions that carry 3-4 marks in the Class 10 Maths Board exam.
Q1.Which term of the AP: 121, 117, 113, . . ., is its first negative term? [Hint: Find n for an < 0]
Solution:
Let the first negative term of the given AP 121, 117, 113, . . ., is an
nth term of a AP is given by
an= a + (n -1)d
Where a = 121, d = 117 -121 =-4
For the first number to be negative
an< 0
a + (n -1)d < 0
121 + (n -1)×-4 <0
(n -1)×-4 < -121
-4n +4 < -121
4n -4 > 121
4n > 121 +4
4n > 125
n > 125/4
n > 31.25
Since n is a natural number therefore first negative term is 32 nd
Q2.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:
The sum of third and seventh terms of an AP is 6
a3+ a7=6
a +2d +a +6d =6
2a +8d =6
a + 4d = 3……(i)
a3× a7=8
(a +2d )(a +6d) = 8
Putting the value of a from the equation (i) a =3 -4d
(3-4d +2d )(3-4d +6d) = 8
(3 -2d )(3 +2d) = 8
9 – 4d² = 8
4d² = 9-8 =1
d² = 1/4
d =±(1/2)
Putting d =1/2 in equation (i)
a + 4×1/2 = 3
a +2 =3
a = 1
Putting d =-1/2 in equation (i)
a + 4×-1/2 = 3
a -2 =3
a = 5
If d=1/2,a=1 then the sum of n terms of the AP is given by
Sn= n/2[2a + (n – 1)d]
=16/2[2×1 + (16 – 1)×1/2]
=8[2 +15/2]
=8×(4+15)/2=4×19 =76
If d=-1/2,a=5 then the sum of n terms of the AP is given by
Sn= n/2[2a + (n – 1)d]
=16/2[2×5 + (16 – 1)×-1/2]
=8[10 -15/2]
=8×(20-15)/2=4×5=20
Q3.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
apart, what is the length of the wood required for the rungs? [Hint: Number of rungs = 250/25 +1].
Solution:
The distance the top and the bottom rungs is
The rungs are 25 cm distance apart
The number of rungs in the ladder
250/25 +1 =10 + 1 =11
Since the rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top,the length between the successive rungs has a common difference therefore the lengths of rungs make an AP.
45,45 +d, 45 +2d+……..25
First term,a =45 and last term,l =25
The sum of the AP shows the length of the wood required for the rungs
Sn=n/2[a +l]
= 11/2[45 +25]
=11/2[70] =11×35 =385
Hence the length of the wood required for the rungs is 385 cm
Q4.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:
It is given that houses are numbered from 1 to 49 and house no. x is 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.
1,2,3……….x-1,x,x+1…………49
Since 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.
Number of the houses preceding the number x are (x -1) and the number of houses following the number x are 49 -x
Given to us
Sx-1=S49-x
The sum of n terms of the AP is given by
Sn=n/2[a +l]
=(x-1)/2[1 +x-1]=x(x -1)/2
S49-x=(49-x)/2 [x+1 +49]=(49-x)/2[x+50]
x(x -1)/2 = (49-x)/2[x+50] =(49x +2450 -x²-50x)/2
x² -x = 49x +2450 -x²-50x
2x² -x +50x -49x – 2450 =0
2x² -2450
2x² = 2450
x² = 2450/2 =1225
x = ±35
Neglecting the negative term because the number of houses can’t be negative, therefore x =35
Q5. 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 = 1/4 ×1/2 ×50 m3.]
Solution: The terrace has 15 steps, each step has a rise of 1 /4 m and a tread of 1 /2 m
The height of the first step, second step, third step up to 15 th step from the ground is
1/4 m, (1/4 m +1/4 m),…………,[1/4+(15-1)1/4]m=1/4 +7/2
⇒1/4 m,1/2 m,3/4 m……,15/4 m
Length × breadth of each step are 50 m ×1/2 m
The volumes of the concrete required for first, second, third up to 15 th step are
(1/4×50 ×1/2 )m³,(1/2×50 ×1/2 )m³,(3/4×50 ×1/2 )m³,…….(15/4×50 ×1/2 )m³
(50/8)m³,(50/4)m³,(150/8)m³……..,(750/8)m³
6.25 m³, 12.5 m³,18.75m³……..,93.75 m³
Since the terms show that it is an AP, therefore total volume of the terrace will be the sum of the AP
The sum of n terms of the AP is given by
Sn=n/2[a +l]
Putting the value n =15,a =6.25, l =93.75
S15=15/2[6.25 +93.75] =15/2[100] =15×50 =750
Hence the total volume of concrete required to build the terrace is 750 m³
The formulas used in this exercise 5.4 are following
nth term of an AP is given by
an is nth term of the AP, d is common difference of the AP, n is total number of terms
Sum of the n terms is given by
Sn is the n terms of the AP
Understanding Class 10 Maths Chapter 5 Exercise 5.4 is key to getting good at Arithmetic Progression (AP). Our easy-to-follow Class 10 Maths Chapter 5 Exercise 5.4 Solutions will guide you through each problem step-by-step. With these solutions, Arithmetic Progression Exercise 5.4 will seem much easier. Be sure to download the PDF for Class 10 Chapter 5 Maths Exercise 5.4 so you can practice anytime and keep these important concepts handy. Keep practicing, and you’ll do great in your exams!
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