Solve the followings Questions.

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

Answer:

Given: 121, 117, 113, …….

Here, a = 121 and d = 117-121 = – 4

Let the nth term of the given A.P. be the first negative term. Then, a_n  < 0.

⇒ 121+ (n – 1) x (- 4) < 0                                    [  a_n = a + (n – 1) d]

⇒ 125 – 4n < 0

⇒ – 4n < -125

⇒

Therefore, n = 32

Hence, the first negative term is 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 sixteen terms of the AP.

Answer:

We know that,

⇒ a_n = a + (n – 1) d

⇒ a₃ = a + (3 – 1) d

⇒ a₃ = a + 2d

Similarly,   a₇ = a + 6d

Given,      a₃ + a₇ = 6

⇒ (a + 2d) + (a + 6d) = 6

⇒ 2a + 8d = 6

⇒ a + 4d = 3

⇒ a = 3 – 4d  (i)

Also, it is given that, (a₃) x (a_{7) = 8}

⇒ (a + 2d)  x (a + 6d) = 8

From Equation (i),

⇒  (3 – 4d + 2d ) x (3 – 4d + 6d) = 8

⇒  (3 – 2d) x (3 + 2d) = 8

⇒   9 – 4d² = 8

⇒   4d² = 9 – 8 = 1

⇒ d² = 1/4

⇒ d = ± 1/2

⇒ d = 1/2 or 1/2

From equation (i)

(When d = 1/2)

⇒ a = 3 – 4d

⇒ a = 3 – 4(1/2)

⇒ 3 – 2 = 1

(When d is – 1/2)

⇒ a = 3 – 4(-1/2)

⇒ a = 3 + 2 = 5

(When a is 1 & d is 1/2)

⇒   8 = [2 + 15/2]

⇒  4 (19) = 76

(When a is 5 & d is  – 1/2)

⇒ 8 [10 + (15(-1/2))]

⇒ 8 (5/2)

⇒ 20

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 5/2 m apart, what is the length of the wood required for the rungs?

Answer:

It is given that the rungs are 25 cm apart and the top and bottom rungs are 2 1/2 m apart.

Therefore,Now, as the lengths of the rungs decrease uniformly, they will be in an A.P.

The length of the wood required for the rungs equals the sum of all the terms of this A.P.

First term, a = 45

Last term, l = 25

n = 11

⇒ S_n = n/2(a+1)

Therefore,S_{10 = 11/2(45+25) = 11/2 + 70 = 385cm.}

_{Therefore, 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.

Answer:

Let there be a value of x  such that the sum of the numbers of the houses preceding the houses numbered x is equal to the sum of the numbers of the houses following it.

That is , 1 + 2 + 3….. + (x + 1) = (x + 1) + (x + 2)………  + 49

Therefore, 1 + 2 + 3….. + (x + 1) = [1 + 2 + ….. + x + (x + 1)…… + 49 ] – [1 + 2 + 3….. + (x)]

Therefore,

⇒  x(x – 1) = 49 x 50 – x(x + 1)

⇒  x(x – 1) + x(x + 1) = 49 x 50

⇒ x² – x + x + x² = 49 x 50

⇒  x² = 49 x 25

Therefore, x = 7 x 5 = 35.

Since, x is not a fraction, the value of x satisfying the given condition exists and is equal to 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 figure). Calculate the total volume of concrete required to build the terrace.

Answer:

Volume of concrete required to build the first step, second step, third step, ……. (in m2) are

From the figure, it can be observed that

1st step is 1/2 m wide,

2nd step is 1 m wide,

3rd step is 3/2 m wide,

Therefore, the width of each step is increasing by 1/2 m each time whereas their height 1/4 m and length

50 m remains the same.

Therefore, the widths of these steps are

1/2,1, 3/2, 2,…

Volume of concrete in 1st step =  

Volume of concrete in 2nd step = 

Volume of concrete in 3rd step = 

It can be observed that the volumes of concrete in these steps are in an A.P.

⇒ a = 25/4

⇒ d= 25/2 – 25/4 = 25/4

and  ,

=

=

= 15/2(100) = 750

Volume of concrete required to build the terrace is 750 m³.

CHAPTER NAME	OLD NCERT	NEW NCERT
Real Numbers	EXERCISE 1.1
	EXERCISE 1.2	1.1	CLICK HERE
	EXERCISE 1.3	1.2	CLICK HERE
	EXERCISE 1.4
Polynomials	EXERCISE 2.1	2.1	CLICK HERE
	EXERCISE 2.2	2.2	CLICK HERE
	EXERCISE 2.3
	EXERCISE 2.4
Pair of Linear Equations in Two Variables	EXERCISE 3.1
	EXERCISE 3.2	3.1	CLICK HERE
	EXERCISE 3.3	3.2	CLICK HERE
	EXERCISE 3.4	3.3	CLICK HERE
	EXERCISE 3.5
	EXERCISE 3.6
	EXERCISE 3.7
Quadratic Equations	EXERCISE 4.1	4.1	CLICK HERE
	EXERCISE 4.2	4.2	CLICK HERE
	EXERCISE 4.3
	EXERCISE 4.4	4.3	CLICK HERE
Arithmetic Progressions	EXERCISE 5.1	5.1	CLICK HERE
	EXERCISE 5.2	5.2	CLICK HERE
	EXERCISE 5.3	5.3	CLICK HERE
	EXERCISE 5.4	5.4 (Optional)	CLICK HERE
Triangles	EXERCISE 6.1	6.1	CLICK HERE
	EXERCISE 6.2	6.2	CLICK HERE
	EXERCISE 6.3	6.3	CLICK HERE
	EXERCISE 6.4
	EXERCISE 6.5
	EXERCISE 6.6
Coordinate Geometry	EXERCISE 7.1	7.1	CLICK HERE
	EXERCISE 7.2	7.2	CLICK HERE
	EXERCISE 7.3
	EXERCISE 7.4
Introduction to Trigonometry	EXERCISE 8.1	8.1	CLICK HERE
	EXERCISE 8.2	8.2	CLICK HERE
	EXERCISE 8.3
	EXERCISE 8.4	8.3	CLICK HERE
Some Applications of Trigonometry	EXERCISE 9.1	9.1	CLICK HERE
Circles	EXERCISE 10.1	10.1	CLICK HERE
	EXERCISE 10.2	10.2	CLICK HERE
Construction
Areas Related to Circles	EXERCISE 12.1
	EXERCISE 12.2	11.1	CLICK HERE
	EXERCISE 12.3
Surface Areas and Volumes	EXERCISE 13.1	12.1	CLICK HERE
	EXERCISE 13.2	12.2	CLICK HERE
	EXERCISE 13.3
	EXERCISE 13.4
	EXERCISE 13.5
Statistics	EXERCISE 14.1	13.1	CLICK HERE
	EXERCISE 14.2	13.2	CLICK HERE
	EXERCISE 14.3	13.3	CLICK HERE
	EXERCISE 14.4
Probability	EXERCISE 15.1	14.1	CLICK HERE
	EXERCISE 15.2