Haryana State Board HBSE 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 Textbook Exercise Questions and Answers.

## Haryana Board 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4

Question 1.

Which term of the AP: 121, 117, 113, ……………… is its first negative term?

Solution:

The given sequence of AP is 121, 117, 113, ……………

Here a = 121

d = a_{2} – a_{1}

= 117 – 121 = – 4

Let nth term of AP be the first negative term,

a_{n} = a + (n – 1)d < 0

a + (n – 1)d < 0

= 121 + (n – 1) Ã— (- 4) < 0

= 121 – 4n + 4 < 0

125 – 4n < 0

125 < 4n 4n > 125

â‡’ n > \(\frac{125}{4}\)

â‡’ n > 31.26

Hence, 32th term of the AP is the first negative term.

Question 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 :

Let a be the first term and d be the common difference of AP. Then,

According to question :

a_{3} + a_{7} = 6

â‡’ a + (3 – 1)d + a + (7 – 1)d = 6

â‡’ a + 2d + a + 6d = 6

â‡’ 2a + 8d = 6

â‡’ a + 4d = 3

â‡’ a = 3 – 4d ……………..(1)

And a_{3} Ã— a_{7} = 8

â‡’ (a + 2d)(a + 6d) = 8

â‡’(3 – 4d+ 2d)(3 – 4d + 6d) = 8

[From equation (1) put a = 3 – 4d]

â‡’ (3 – 2d) (3 + 2d) = 8

â‡’ 9 – 4d^{2} = 8

â‡’ 4d^{2} = 9 – 8

â‡’ 4d^{2} = 1

â‡’ d^{2} = \(\frac{1}{4}\)

d = Â± \(\frac{1}{2}\)

Case I:

If d = \(\frac{1}{2}\)

Putting the value of d in equation (1), we get

a = 3 – 4 Ã— \(\frac{1}{2}\)

a = 3 – 2 = 1

a = 1

Now S_{16} = \(\frac{16}{2}\) [2 Ã— 1 + (16 – 1) Ã— \(\frac{1}{2}\)]

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

S_{16} = 8 Ã— \(\frac{19}{2}\)

â‡’ S_{16} = 76.

Case II:

If d = – \(\frac{1}{2}\)

Putting the value of d in equation (1), we get

a = 3 – 4 Ã— (- \(\frac{1}{2}\))

â‡’ a = 3 + 2 = 5

Now, S_{16} = \(\frac{16}{2}\) [2 Ã— 5 + (16 – 1) Ã—(- \(\frac{1}{2}\))]

â‡’ S_{16} = 8 [10 – \(\frac{15}{2}\)]

S_{16} = 8 Ã— \(\frac{5}{2}\)

â‡’ S_{16} = 20

Hence, S_{16} = 76 or S_{16} = 20.

Question 3.

A ladder has rungs 25 cm apart (See figure). 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 2 \(\frac{1}{2}\) m apart, what is the length of the wood required for the rungs?

Solution:

We have,

Gap between two consecutive rungs = 25 cm

Distance between bottom and top rungs = 2\(\frac{1}{2}\) m = 250 cm

Number of rungs = \(\frac{250}{25}\) + 1 = 11

Since, rungs are decreasing uniformly in length from 45 cm at the bottom to 25 cm at the top.

Therefore, the lengths of rungs form an AP in which a = 45, a_{11} = 25

Required length of wood for rungs = S_{11}

= \(\frac{11}{2}\) (45 + 25)

[âˆµ S_{n} = \(\frac{n}{2}\) (a + l)]

= \(\frac{11}{2}\) Ã— 70

= 385 cm

or = 3.85 m

Hence, required length of wood for rungs = 3.85 m.

Question 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 sum of numbers of the houses following it. Find this value of x.

Solution:

Let there be 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 numÂ¬bers of the houses following it i.e. :

House No. 1, 2, 3, …………….., (x – 1), x, (x + 1), ……………, 49.

Now, S_{x – 1} = S_{49} – S_{x}

x^{2} = 1225

x = âˆš 1225

x = Â± 35.

Since, x canâ€™t be negative. Therefore, x = 35.

Question 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 \(\frac{1}{4}\) m and a tread of \(\frac{1}{2}\) m (see figure). Calculate the total volume of concrete required to build the terrace.

Solution:

We have, Length and breadth of each step are 50 m and \(\frac{1}{2}\) m respectively.

Height of first step = \(\frac{1}{4}\) m

Height of second step = \(\frac{1}{4}+\frac{1}{4}=\frac{2}{4}\) m

Height of third step = \(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\) m

and so on.

Volume of concrete to build the first step = 50 Ã— \(\frac{1}{2} \times \frac{1}{4}=\frac{25}{4}\) m^{3}

[âˆµ Volume of cuboidal step = L Ã— B Ã— H]

Volume of concrete to build the second step = 50 Ã— \(\frac{1}{2} \times \frac{1}{4}=\frac{50}{4}\) = m^{3}

Volume of concrete to build the third step = 50 Ã— \(\frac{1}{2} \times \frac{3}{4}=\frac{75}{4}\) m^{3}

and so on.

The sequence of volume of concrete 0f steps are: \(\frac{25}{4}, \frac{50}{4}, \frac{75}{4}, \ldots\)

a_{2} – a_{1} = \(\frac{50}{4}-\frac{25}{4}=\frac{25}{4}\)

a_{3} – a_{2} = \(\frac{75}{4}-\frac{50}{4}=\frac{25}{4}\)

âˆµ a_{2} – a_{1} = a_{3} – a_{2}

âˆ´ The sequence forms an AP in which a = \(\frac{25}{4}\) and d = \(\frac{25}{4}\).

Now, total volume of concrete = Sum of volumes of concrete ta build 15 steps

= S_{15}

= \(\frac{15}{2}\left[2 \times \frac{25}{4}+(15-1) \times \frac{25}{4}\right]\)

= \(\frac{15}{2}\left[\frac{25}{2}+14 \times \frac{25}{4}\right]\)

= \(\frac{15}{2}\left[\frac{25}{2}+\frac{175}{2}\right]\)

= \(\frac{15}{2} \times \frac{200}{2}\)

= 15 Ã— 50

= 750 m^{3}

Hence, the total volume of concrete = 750 m^{3}.