ex 6 5 fsc part1

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Exercise 6.5 (Solutions)

Textbook of Algebra and Trigonometry for Class XIMerging man and maths Available online @ http://www.mathcity.org, Version: 1.0.0

Question # 1

The sequence of the deposits is10, 15, 20, ……………. to 9 terms

Here a1 =10 , d = 15 − 10 = 5 , n = 9Since

S = n

[2a + (n −1)d ]2

1

⇒ S9

= 9

[2(10) + (9 −1)(5)] =

9 [20 + 40]

=9

(60) = 2702 2 2

Hence the total amount he deposits is Rs. 270.

Question # 2The sequence of the trees from top to base row

is 1, 2, 3, ……………Let n be the total number of tress in base row then

a1 = 1, d = 2 − 1 =1 , n = n , Sn = 378

Now Sn

= n

[2a + (n

−1)d ]2

1

⇒ 378 = n

[2(1) + (n −1)(1)] ⇒ 756 = n [2 + n

−1]2

⇒ 756 = n (n +1) ⇒ 756 = n 2 + n

1Þ n 2 + n − 756 = 0

Son =

−1 ±

( − 1) 2 − 4(1)( − 756)

2(1)

⇒ n = − 1

±1 + 3024

= − 1

±2

3025 =

− 1 ± 55

2 2

So

n = − 1 + 55

=54 = 27 or

n = − 1 − 55

=−

56= − 28

2 2 2 2Since n can never be negative therefore n = 27

Now an = a1 + (n −1)d

⇒ a27 = (1) + (27 −1)(1) = 1+ 26 = 27Thus the numbers of trees in the base row are 27.

Question # 3Let the first installment be x then the sequence of installment will be

x, x − 10, x − 20,......................

Here a1 = x ,d = -10 , n = 14 and Sn = 1100 + 230 = 1330

Now Sn

= n

[2a + (n −1)d ]2

1

⇒ 1330

=

14 [2x + (14 − 1)(−10)] ⇒ 1330 = 7[2x −130] ⇒ 1330 = 14x −910

2

2240 = 160⇒ 1330+ 910 = 14x ⇒ 2240 = 14x ⇒ x =

14Hence the first installment is 160.

FSc-I / Ex 6.5 - 2

Question # 4The sequence of the strikes is

1, 2, 3, …………., 12

Herea1 = 1

, d = 2 – 1 = 2 , n = 12 , an = 12

Now Sn

= n

[2a + (n −1)d

]2

1

⇒

S

n

= 12[2(1) + (12 −1)(1)]

=12

[2 + 11] =

12

[13] = 78

2 2 2Hence clock strikes 78 hours in twelve strikes.

Question # 5The sequence of the savings is

12, 16, 20, ……………Total Savings = 2100So here a1 = 12 , d = 16 – 12 = 4 , Sn = 2100 , n = ?

Since S

= n

[2a + (n −1)d

]2

1

⇒ 2100 = n

[2(12) + (n −1)(4)] ⇒ 4200 = n [24 + 4 n − 4]2

⇒ 4200 = n [4n + 20]⇒ 4200 = 4n 2

+ 20n

⇒

4n 2 + 20n − 4200 = 0 ⇒

4(n 2 + 5n − 1050) = 0

⇒ n 2 + 5n − 1050 = 0

−5±

(5)2 − 4(1)(−1050) − 5 ±

⇒ n = =

25 + 4200

2(1) 2

− 5 ±

= − 5 ±

65=4225

2 2

So n = − 5 − 65 =−70

= −35 orn =

− 5 + 65

=

60

=30

2 2 2 2As n can never be negative therefore n = 30

Thus student will save Rs. 2100 in 30 weeks.

Question # 6The sequence of the falls is

9, 27, 45, ………………(1) a1 = 9 , d = 27 – 9 = 18 , a5 = ?

Since a5 = a1 + 4d = 9 + 4(18) = 9 + 72 = 81

Hence in fifth second the object will fall 81 meters.(2) Here a1 = 9 , d = 27 – 9 = 18 , n = 5, S5 = ?

Since Sn = n2 [2a1 + (n −1)d ]

1Þ S5 = 52 [2(9) + (5 −1)(18)] =

52 [18 + 72] =

52 (90) =

225 Thus up to 5th second the object will fall 225 meters.

FSc-I / Ex 6.5 - 3

Question # 7

Here a1 = 6000 , a11 = 12000 , n = 11

Now Sn

= n

[a1 + an

]2

⇒ S = 11[6000 +12000] =

11(18000) = 99000

11

2 2Hence he will receive Rs. 99000 in past eleven years.

Question # 8Since the sum of angels of 3 sided polygon (triangle) = a1 = π

Sum of angels of 4 sided polygon (quadrilateral) = a2 = 2πSum of the angels of 5 sided polygon (pentagon) = a3 = 3π

SoThe sum of interior angels of 16 side polygon = a14 = ?

Here a1 = π , d = a2 − a1 = 2π − π = π , n = 14

Since an = a1 + (n −1) d

⇒ a14 = π + (14 −1)(π ) = π + 13π = 14πHence sum of interior angels of 16 side polygon is 14π .

Question # 9

Let a1 denotes the prize money for the last positionThen a1 = 4000 , Sn = 60000 , n = 8 , an = ?

Since S

= n (a +

a )2 1

⇒ 60000 = 8

(4000+

a ) ⇒ 60000 = 4(4000+ a )2 n

⇒ 60000 = 16000 + 4an

⇒ 60000 − 16000 = 4an ⇒ 44000 = 4an

⇒ an = 44000

=

110004

Hence the team at 1st place will get 11000 Rs.

Question # 10Balls in the first layer = 8 + 7 + 6 +……… + 2 + 1

= 8

[2(8) + (8 −1)(−1)] = 4(16 − 7)

=36

2Balls in the second layer = 7 + 6 + 5 + ………+ 2 + 1

= 7

[2(7) + (7 −1)(−1)] = 7

[14− 6 ]= 7 [ 8 ]=28

2 2 2Balls in the third layer

= 6 + 5 + 4 +3 + 2 + 1 = 21

Balls in the fourth layer = 5 + 4 + 3 + 2 +1 = 15Balls in the fifth layer = 4 + 3 + 2 + 1 = 10Balls in the sixth layer = 3 + 2 + 1 = 6Balls in the seventh layer = 2 + 1 = 3Balls in the eighth layer = 1Hence the number of balls in pyramid

= 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 120

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