NCERT

10th Mathematics

Exercise 5.4

Topic: 5 Arithmetic Progressions

### 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:

1) According to the problem, a_{1}_{ }= 121_{, }a_{2}_{ }= 117_{, }a_{3}_{ }= 113. . . so d = - 4,

2) We will have to find n using the above information.

a_{n} = a + (n – 1) d

a_{n} = 121 + (n – 1) (- 4)

a_{n} = 121 + (- 4n + 4)

a_{n} = 121 - 4n + 4

a_{n} = 125 - 4n

3) We have to find which first term is negative.

n > 31.25

4) So 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 a_{3} + a_{7} = 6 ---------- equation 1

3) Here a_{3} x a_{7} = 8 ---------- equation 2

4) Now we will find a_{3} and a_{7},

a) First we will find a_{3}

a_{3} = a + (3 – 1) d

a_{3} = a + 2d ---------- equation 3

b) Now we will find a_{7}

a_{n} = a + (n – 1) d

a_{7} = a + (7 – 1) d

a_{7} = a + 6d ---------- equation 4

5) From equations 1, 3, and 4, we have,

a_{3} + a_{7} = 6

a + 2d + a + 6d = 6

2a + 8d = 6

2(a + 4d) = 6

(a + 4d) = 6/2

(a + 4d) = 3

a = 3 - 4d ---------- equation 5

6) From equations 2, 3, 4, and 5, we have,a_{3} x a_{7} = 8(a + 2d) x (a + 6d) = 8

(3 - 4d + 2d) x (3 - 4d + 6d) = 8

(3 - 2d) x (3 + 2d) = 8

(3^{2} - 4d^{2}) = 8

4d^{2} = 9 - 8

4d^{2} = 1

d^{2} = 1/4

d^{2} = 1/4

d = ± 1/2

d = 1/2 or - 1/2 ---------- equation 6

7) Put d = 1/2 and d = - 1/2 from equation 6 in equation 5,a) First we take d = 1/2, we get

a = 1 ---------- equation 7

b) First we take d = - 1/2, we get

a = 3 + 2

a = 5 ---------- equation 8

8) We know that the sum of the first n terms of an AP is given by:

S_{n} = (n/2)[2a + (n - 1) d]

a) First we will find sum of first 16 terms with a = 1 and d = 1/2:

S_{n} = (n/2)[2a + (n - 1) d]

S_{16} = (16/2)[2(1) + (16 - 1) (1/2)]

S_{16} = 8[2 + (15) (1/2)]

S_{16} = 8[2 + (15/2)]

S_{16} = 8[(4 + 15)]/2

S_{16} = 4(19)

S_{16} = 76 ---------- equation 9

b) First we will find sum of first 16 terms with a = 5 and d = (- 1/2):

S_{n} = (n/2)[2a + (n - 1) d]

S_{16} = (16/2)[2(5) + (16 - 1) (- 1/2)]

S_{16} = 8[10 + (15) (- 1/2)]

S_{16} = 8[10 - (15/2)]

S_{16} = 8[(20 - 15)]/2

S_{16} = 4(5)

S_{16} = 20 ---------- equation 10

9) From equations 9 and 10, we have,

a) S_{16} = 76 when a = 1 and d = 1/2

b) S_{16} = 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, a_{1} = 45, l = 25 and n = 11, so,

S_{n} = (n/2)[a + l]

S_{11} = (11/2)[45 + 25]

S_{11} = (11/2)[70]

S_{11} = 11(35)

S_{11} = 385

6) 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 : s**_{x - 1} = S_{49} – S_{x}]

### Solution:

1) Row houses are numbered 1, 2, 3, . . 49.

2) So, a_{1}_{ }= 1_{, }a_{2}_{ }= 2_{, }a_{3}_{ }= 3. . . so d = 1,

3) The sum of the number of houses preceding the xth house will be S_{(x-1)}.

S_{n} = (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

4) Now we will find S_{49}S_{n} = (n/2)[2a + (n – 1) d]S_{49} = (49/2)[2(1) + (49 - 1) (1)]

S_{49} = (49/2)(2)[1 + 24]

S_{49} = 1225 ---------- equation 2

5) Now we will find S_{x}

S_{n} = (n/2)[2a + (n – 1) d]S_{x} = (x/2)[2(1) + (x - 1) (1)]

S_{x} = (x/2)[2 + (x - 1)]

S_{x} = (x/2)[x + 1]S_{x} = x(x + 1)/2 ---------- equation 3

6) Now we will find S_{49} – S_{x }using equations 2 and 3.

S_{49} - S_{x} = 1225 - [x(x + 1)/2] ---------- equation 4

7) According to the problem, S_{(x-1)} = S_{49} - S_{x } ---------- equation 5

8) From equations 1, 4, and 5, we have,

S_{(x-1)} = S_{49} - S_{x}

x(x-1)/2 = 1225 - [x(x + 1)/2]

x(x-1)/2 = [2450 - (x(x + 1))]/2

x(x-1) = [2450 - (x(x + 1))]

x^{2 }- x = 2450 - x^{2 }- x

2x^{2 }= 2450

x^{2 }= 1225

x^{ }= ± 35

x^{ }= 35 or - 35

9) As 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 m**^{3}]

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) m

b) The height of the second step is (1/4) + (1/4) = (1/2) m

c) The height of the 3rd step is (1/2) + (1/4) = (3/4) m

d) 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,

S_{n} = (n/2)[2a + (n – 1) d]

S_{15} = (15/2)[2(25/4) + (15 – 1) (25/4)]

S_{15} = (15/2)[2(25/4) + (14) (25/4)]

S_{15} = (15/2) x (2(25/4)) x [1 + (7)]

S_{15} = (15/2) x (2(25/4)) x [8]

S_{15} = [(15 x 2 x 25)/8] x [8]

S_{15} = (15 x 2 x 25)

S_{15} = (30 x 25)

S_{15} = (750)

9) The concrete required to build the terrace is 750 m^{3}.

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