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# NCERT Solutions for Class 10 Maths Exercise 5.4 Chapter 5 Arithmetic Progression

NCERT Solutions for Class 10 Maths Exercise 5.4 Chapter 5 Arithmetic Progression are presented here to boost preparation for the CBSE exam. NCERT Solutions for Class 10 Maths Exercise 5.4 Chapter 5 Arithmetic Progression are created here to boost preparation for the CBSE exam.NCERT Solutions for Class 10 Maths Exercise 5.4 Chapter 5 Arithmetic Progression is the best study material for the doubt clearance on chapter  5 – Arithmetic Progression. NCERT Solutions for Class 10 Maths Exercise 5.4 Chapter 5 Arithmetic Progression are valuable in anticipating the actual quantity in the field of financial matters, trade, science, and other problems when quantities are in a sequence.NCERT Solutions for Class 10 Maths Exercise 5.4 of Chapter 5 Arithmetic Progression are solved by an experienced CBSE maths teacher according to the standards of CBSE. All questions of exercise 5.4 are solved here by a step-by-step technique so every class 10 maths student can comprehend the solutions without any problem.

Class 10 Maths  Chapter 5 Arithmetic Progression contains four exercises, exercise 5.1, exercise 5.2, exercise 5.3, and exercise 5.4, here links of each exercise are given, so you can study NCERT solutions of whole Chapter 5 Arithmetic Progression.

Why are NCERT Solutions for Class 10 Maths Exercise 5.4 Chapter 5 Arithmetic Progression important for you?

• All unsolved Questions of the exercise Exercise 5.4 are covered.
• All questions are solved in detail by a step by step way.
• These NCERT Solutions gives an idea about the way of solving the similar questions.
• Exercise 5.4 covers the questions of 3-4 marks in class 10 maths board exam.

## NCERT Solutions for Class 10 Maths Exercise 5.4 Chapter 5 Arithmetic Progression

Q1.Which term of the AP: 121, 117, 113, . . ., is its first negative term? [Hint: Find n for an < 0]

Ans. 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.

Ans.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].

Ans.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

### NCERT Solutions for Class 10 Maths Exercise 5.4 Chapter 5 Arithmetic Progression

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 ]

Ans.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.]

Ans. 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

We hope you would have liked the solutions of the class 10 maths exercise 5.4 Arithmetic Progression,the questions of exercise 5.4 are very important not only for the preparation of class 10 maths paper in term 2 CBSE Board exam but also useful in the preparation of competitive entrance exams. If you liked the solutions then please don’t forget to subscribe share our website for updated information because without your support we can’t survive on the web.

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