Gujarat Board GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 1.

Find the sum of the following APs

(i) 2, 7, 12, … to 10 terms

(ii) – 37, – 33, – 29, ……… to 12 terms

(iii) 0.6, 1.7, 2.8, … to 100 terms

(iv) \(\frac { 1 }{ 15 } \), \(\frac { 1 }{ 12 } \), \(\frac { 1 }{ 10 } \)to 11 terms.

Solution:

(i) 2, 7, 12, … to 10 terms

Here a = 2, d = 7 – 2 = 5

We know that

S_{n} = \(\frac { n }{ 2 } \) [2a + (n – 1)d]

S_{10} = \(\frac { 10 }{ 2 } \) [2 x 2 + (10 – 1) x 5 ]

S_{10} = 5 [4 + 9 x 5 ]

S_{10} = 5 x 49

S_{10} = 245

Hence the sum of 10 terms of AP is 245.

(ii) – 37, – 33, – 29, … to 12 terms

Here a = – 37, d = – 33 + 37 = 4

We know that sum of AP is

S_{n}= \(\frac { n }{ 2 } \) [2a + (n – 1)d]

S_{12} = \(\frac { 12 }{ 2 } \) [2 x ( – 37) + (12 – 1) x 4]

S_{12} = 6 [ – 74 + 44]

S_{12} = 6 x – 30

S_{12} = -180

Hence the sum of 12 terms of AP is – 180.

(iii) 0.6, 1.7, 2.8, … to 100 terms

Here a = 0.6 and d = 1.7 – 0.6 = 1.1

We know the sum of AP is

S_{n} = \(\frac { n }{ 2 } \) [2a + (n – 1) d]

S_{100} = \(\frac { 100 }{ 2 } \) [2 x 0.6 + (100 – 1) x 1.1]

S_{100} = 50 [1.2 + 99 x 1.1]

S_{100} = 50 [1.2 + 108.9]

S_{100} = 50 [110.1]

S_{100} = 5505

The um of loo terms of AP is 5505.

(iv) \(\frac { 1 }{ 15 } \), \(\frac { 1 }{ 12 } \), \(\frac { 1 }{ 10 } \), ……..to 11 terms

Here a = \(\frac { 1 }{ 15 } \)

and d = \(\frac { 1 }{ 12 } \) – \(\frac { 1 }{ 15 } \) = \(\frac { 5-4 }{ 60 } \)

d = \(\frac { 1 }{ 60 } \)

n = 11

We know that the sum of AP is

S_{n} = [2a + (n – 1)d]

S_{11} = \(\frac { 11 }{ 1 } \) \(\left[ 2x\frac { 1 }{ 15 } +(11-1)x\frac { 1 }{ 60 } \right] \)

S_{11} = \(\frac { 11 }{ 2 } \) \(\left[ \frac { 2 }{ 15 } +\frac { 10 }{ 60 } \right] \)

S_{11} = \(\frac { 11 }{ 2 } \) \(\left[ \frac { 2+10 }{ 60 } \right] \)

S_{11} = \(\frac { 11 }{ 2 } \) x \(\frac { 18 }{ 60 } \)

S_{11} = \(\frac { 33 }{ 20 } \)

Sum of 11 terms of AP is \(\frac { 33 }{ 20 } \).

Question 2.

Find the sums given below:

(i) 7 + 10\(\frac { 1 }{ 2 } \) + 14 + ……… + 84

(ii) 34 + 32 + 30 + ……….. + 10

(iii) – 5 + (-8) + (-11) + ………… + (-230)

Solution:

(i) 7 + 10\(\frac { 1 }{ 2 } \) + 14 + ……… + 84

Here

a = 7, d = 10\(\frac { 1 }{ 2 } \) – 7

d = \(\frac { 21 }{ 2 } \) – 7 = \(\frac { 7 }{ 2 } \)

a_{n} = a + (n – 1) d

84 = 7 + (n – 1) \(\frac { 7 }{ 2 } \)

77 = (n – 1) \(\frac { 7 }{ 2 } \)

n – 1 = \(\frac { 77 x 2 }{ 7 } \)

n – 1 = 22

n = 23

Sum of AP, S_{n} = \(\frac { n }{ 2 } \) \(\frac { 1 }{ 2 } \)

S_{23} = \(\frac { 23 }{ 2 } \) \(\left[ 2×7+(23-1)x\frac { 7 }{ 2 } \right] \)

S_{23} = \(\frac { 23 }{ 2 } \) \(\left[ 14+22x\frac { 7 }{ 2 } \right] \)

S_{23} = \(\frac { 23 }{ 2 } \) \(\left[ 14+77 \right] \)

S_{23} = \(\frac { 23 }{ 2 } \) x 91

S_{23} = \(\frac { 2093 }{ 2 } \)

S_{23} = 1046\(\frac { 1 }{ 2 } \)

Hence the required sum is 1046\(\frac { 1 }{ 2 } \)

(ii) 34 + 32 + 30 + ………. + 10

Here a = 34, d = 32 – 34 = – 2

l = 10

Let the number of terms of the AP be n.

Then l = a + (n – 1)d

10 = 34 + (n – 1) x ( – 2)

\(\frac { -24 }{ -2 } \) = n – 1

n – 1 = 12

n = 13

We know that

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

S_{13} = \(\frac { 13 }{ 2 } \) [34 + 10]

S_{13} = \(\frac { 13 }{ 2 } \) x 44

S_{13} = 13 x 22

S_{13} = 286

Hence required sum is 286.

(iii) -5 + (-8) + (-11) + ……….. + (-230)

Here a = – 5, d = – 8 – (- 5)

d = – 8 + 5 = – 3

l = – 230

Let n be the number of terms is AP. Then,

l = a + (n – 1)d

-230 = – 5 + (n – 1) (- 3)

(n – 1) x (- 3) = – 230 + 5

n – 1 = \(\frac { 225 }{ 3 } \) = 75

n = 76

The sum of AP is

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

S_{76} = \(\frac { 76 }{ 2 } \) [ – 5 + (- 230)]

S_{76} = \(\frac { 76 }{ 2 } \) [ – 5 – 230]

S_{76} = \(\frac { 76 }{ 2 } \) [- 235]

S_{76} = 38 x -235

S_{76} = – 8930

Hence the required sum is -8930.

Question 3.

In an AP:

(i) Given a = 5, d = 3, a_{n} = 50, find n and S_{n}.

(ii) Given a = 7,a_{13} = 35. find d and S_{13}.

(iii) Given 12 = 37, d = 3, find a and S_{12}.

(iv) Given a = 15, S_{10} = 125, find d and S_{12}.

(v) Given d = 5, S_{9} = 75, find a and a_{9}.

(vi) Given a = 2, d = 8, S_{n} = 90, find n and a_{n}.

(vii) Given a = 8, a_{n} = 62, S_{n} = 210 find n and d.

(viii) Given a_{n} = 4, d = 2, S_{n} = – 14, find n and a.

(ix) Given a = 3, n = 8, S = 192 fInd d.

(x) Given l = 28, S = 144, and thora nra total 9

terms, find a

Solution:

(i) Here a = 5

d = 3

a_{n} = 50

We know that

a_{n} = a + (n – 1) d

50 = 5 + (n – 1) x 3

n – 1 = \(\frac { 45 }{ 3 } \)

n – 1 = 15 + 1

n = 16

Sum of AP is

S_{n} = \(\frac { n }{ 2 } \) [2a + (n – 1) d]

S_{n} = \(\frac { -16 }{ 2 } \) [2 x 5 + (16 – 1) x 3 ]

S_{n} = 8 [ 10 + 45 ]

S_{n} = 8 x 55

S_{n} = 440

(ii) Here

a = 7

a_{13} = 35

a + 12d = 35

7 + 12d = 35

12d = 28

d = \(\frac { 28 }{ 12 } \) = \(\frac { 7 }{ 3 } \)

d = \(\frac { 7 }{ 3 } \)

We know that

S_{n} = \(\frac { n }{ 2 } \) [2a + (n – 1)d]

S_{13} = \(\frac { 13 }{ 2 } \) [ 2 x 7 + (13 – 1) x \(\frac { 7 }{ 3 } \)]

S_{13} = \(\frac { 13 }{ 2 } \) [ 14 + 12 x \(\frac { 7 }{ 3 } \) ]

S_{13} = \(\frac { 13 }{ 2 } \) [14 + 28]

S_{13} = \(\frac { 13 }{ 2 } \) x 42

S_{13} = 13 x 21

S_{13} = 273

(iii) Here a_{12} = 37

Now, d = 3

a_{12} = 37

a + 11d = 37

a + 11 x 3 = 37

a = 37 – 33

a = 4

s_{12} = \(\frac { 17 }{ 2 } \) [ 2a + (n – 1)d]

s_{12} = \(\frac { 12 }{ 2 } \) [ 2 x 4 + ( 12 – 1) x 3]

s_{12} = 6 [ 8 + 11 x 3]

s_{12} = 6 x 41

s_{12} = 246

(iv) Here a_{3} = 15

Now, S_{10} = 125

a_{3} = 15

a + 2d = 15

and S_{n} = \(\frac { n }{ 2 } \) [ 2a + (n – 1) d]

S_{10} = \(\frac { 10 }{ 2 } \) [ 2 x a + (10 – 1) x d]

125 = 5 [ 2a + 9d]

25 = 2a + 9d

2a + 9d = 25

Solving eqn. (1) and (2)

Putting value of d = – 1 in eqn. (1)

a + 24 = 15

a + 2 x (- 1) = 15

a – 2 = 15

a = 15 + 2

a = 17

them a_{10} = a + 9d

a_{10} = 17 + 9 x ( – 1)

a_{10} = 17 – 9

(v) Here d = 5

S_{9} = 75

We know that

S_{n} = \(\frac { n }{ 2 } \) [2a + (n – 1)d]

S_{9} = \(\frac { 9 }{ 2 } \) [2 x a + (9 – 1) x 5]

75 = \(\frac { 9 }{ 2 } \) [ 2a + 8 x 5]

150 = 9 x [2a + 40]

2a = \(\frac { 50 }{ 3 } \) – 40

2a = \(\frac { 50 – 120 }{ 3 } \)

2a = \(\frac { -70 }{ 3 } \)

a = \(\frac { -70 }{ 3×2 } \)

a = \(\frac { -35 }{ 3 } \)

Again we know that

a_{n} = a + (n – 1) d

a_{9} = a + (9 – 1)d

a_{9} = a + 8d

a_{9} = \(\frac { -35 }{ 3 } \) + 8 x 5

a_{9} = \(\frac { -35 }{ 3 } \) + 40

a_{9} = \(\frac { -35+120 }{ 3 } \)

a_{9} = \(\frac { 85 }{ 3 } \)

(vi) Here a = 2, d = 8,

and S_{n} = 90

We know that

S_{n} = \(\frac { n }{ 2 } \) [ 2a + (n – 1) x 8]

90 = \(\frac { n }{ 2 } \) [2 x 2 + (n – 1) x 8]

180 = n [4 + 8n – 8]

180 = n [8n – 4]

180 = 4n [2n – 1]

\(\frac { 180 }{ 4 } \) = 2n^{2} – n

45 = 2n^{2} – n

2n^{2} – n – 45 = 0

2n(n – 5) + 9(n – 5) = 0

(n – 5)(2n + 9) = 0

n = 5 or n = \(\frac { -9 }{ 2 } \)

It is number of term. So it will be a natural number. Hence, n = 5

(vii) Here a = 8, a_{n} = 62, S_{n} = 210

We know that

a_{n} = a + (n – 1)d

62 = 8 + (n – 1) x d

54 = ( n – 1) d

(n – 1) d = 54 …………… (1)

S_{n} = \(\frac { n }{ 2 } \) [2a + (n – 1)d]

210 = \(\frac { n }{ 2 } \) [2 x 8 + (n – 1) d] …………(2)

From eqn. (1) and (2) we get

210 = \(\frac { n }{ 2 } \) [16 + 54]

210 = \(\frac { n }{ 2 } \) x [70]

n = \(\frac { 210 }{ 35 } \)

n = 6

Putting n = 6 in eqn. (1), we get(n – 1)d = 54

(6 – 1)d = 54

5d = 54

d = \(\frac { 54 }{ 5 } \)

(viii) Herr a_{n} = 4

d = 2

S_{n} = – 14

We know that

a_{n} = a + (n – 1)d

4 = a + (n – 1) 2

4 = a + 2n – 2

a + 2n = 6

Now S_{n} = – 14

We know that

S_{n} = \(\frac { n }{ 2 } \)[2a + (n – 1)d]

– 14 = \(\frac { n }{ 2 } \) [2a + (n – 1) 2]

– 14 = \(\frac { n }{ 2 } \) [a + n – 1]

– 14 = n[6 – n – 1]

[From eqn. (1) a + 2n = 6 â‡’ a + n = 6 – n]

– 14 = n[5 – n]

– 14 = 5n – n^{2}

n^{2} – 5n – 14 = 0

n^{2} – 7n + 2n – 14 = 0

n^{2} – 7n + 2(n – 7) = 0

n(n – 7) + 2(n – 2) = 0

(n – 7) (n + 2) = 0

n = 7 or n = – 2

n = – 2 is impossible because n is a number of terms.

Therefore, n = 7

Putting value of in eqn. (1)

a + 2n = 6

a +2 x 7 = 6

a = 6 – 14

a = – 8

(ix) It is given that

a = 3

n = 8

S = 192

We know that

8 = \(\frac { n }{ 2 } \) [2a + (n – 1)d]

192 = \(\frac { 8 }{ 2 } \)[2 x 3 + (8 – 1) x d]

\(\frac { 192 }{ 4 } \) = [16 + 7d]

48 = 6 + 7d

7d = 42

d = 6

(x) It is given that

l = 28

S = 144

n = 9

We know that

S = \(\frac { n }{ 2 } \) [a + l]

144 = \(\frac { 9 }{ 2 } \) [a + 28]

\(\frac { 144×2 }{ 9 } \) = a + 28

32 = a + 28

a = 4

Question 4.

How many terms of the AP: 9, 17., 25, ……………. must be taken to give s sum of 636?

Solution:

Given AP is 9, 17, 25, …………..

Here a = 9, d = 17 – 9 = 8

Let n terms must be taken, then

S_{n} = 636

S_{n} = \(\frac { n }{ 2 } \) [2a + (n – 1) d]

\(\frac { n }{ 2 } \)[2a x 9 + (n – 1) x 8] = 636

n[9 + 4n – 4] = 636

n[4n + 5] = 636

4n^{2} + 5n = 636

4n^{2} + 5n – 636 = 0

4n^{2} + 53n – 48n – 636 = 0

n(4n + 53) – 12(4n + 53) = 0

(4n + 53)(n – 12) = 0

4n + 53 = 0 or n – 12 = 0

n = \(\frac { -53 }{ 4 } \) (Rejected) or n = 12

Therefore. 12 terms of the AP must be taken.

Question 5.

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Solution:

It is given that

a = 5

l = 45

S = 400

We know that

S = \(\frac { n }{ 2 } \) [a + l]

400 = \(\frac { n }{ 2 } \) [5 + 45]

800 = n [50]

n = \(\frac { 800 }{ 50 } \) = 16

Hence the number of terms is 16. We also know

that

l = a + (n – 1 )d

45 = 5 + (16 – 1) x d

40 = 15d

d = \(\frac { 40 }{ 15 } \)

d = \(\frac { 8 }{ 3 } \)

Hence the common difference is \(\frac { 8 }{ 3 } \)

Question 6.

The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9. how many terms are there and what is their sum?

Solution:

Here a = 17

l = 350

d = 9

We know that

l = a + (n – 1)d

350 = 17 + (n – l) x 9

350 – 17 = (n – 1)9

(n – 1) = \(\frac { 333 }{ 9 } \)

n – l = 37

n = 38

So, there are 38 terms. We know that

S_{n}= \(\frac { n }{ 2 } \) (a.I)

S_{38} = \(\frac { 38 }{ 2 } \) (17 + 350)

S_{38}= 19 x 367

S_{38} = 6973

Hence sum is 6973.

Question 7.

Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Solution:

Here d = 7

a_{22} = 149

We know that

a_{n} = a + (n – 1)d

a_{22} = a + (22 – 1)d

149 = a + 21 x 7

a = 149 – 147

a = 2

Sum of first 22 terms

S_{22} = \(\frac { n }{ 2 } \) [2a + (n – 1)d]

S_{22} = \(\frac { 22 }{ 2 } \) [2 x 2 + (22 – 1) x 7]

S_{22} = 11[4 + 21 x 7]

S_{22} = 11 [4 + 147]

S_{22} = 11 x 151

S_{22} = 1661

Hence sum of 22 terms is 1661.

Question 8.

Find the first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution:

Let a and d are the first term and the common difference respectively.

Second term = 14

(a_{n} = a + (n – 1) d)

a + d = 14 …………… (1)

Third term = 18

a + 2d = 18 ……………(2)

Solving eqn. (1) and (2)

a = 10 and d = 4

Now sum of first 51 terms of the AP

S_{51} = [2a + (51 – 1) d]

(S_{n} = \(\frac { n }{ 2 } \)[2a + (n – 1) d])

S_{51}= \(\frac { 51 }{ 2 } \) [2a + 50 d]

S_{51} = 51 [a + 25d]

= 51 [10 + 25 x 4 ]

= 51 [10 + 100]

= 51 x 110

= 5610

Question 9.

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Solution:

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

Sum of first seven terms = 49

S_{7} = 49

(S_{n} = \(\frac { n }{ 2 } \)[2a + (n – 1) d])

= \(\frac { 7 }{ 2 } \) [2a + (7 – 1)d] = 49

\(\frac { 17 }{ 2 } \) [2a + 6d] = 49

a + 3d = 7 …………. (1)

Sum of first 17 terms – 289

S_{17} = 289

\(\frac { 17 }{ 2 } \) [ 2a + (17 – 1)d] = 289

17 [a + 8d] = 289

a + 8d = \(\frac { 289 }{ 17 } \)

a + 8d = 17 ………….. (2)

Solving eqn (1) and (2), we get

a = 1 and d = 2

Sum of first n terms

S_{n} = \(\frac { n }{ 2 } \) [2a + (n – 1)d]

= \(\frac { n }{ 2 } \) [2 x 1 + (n – 1) x 2]

=n [1 + n – 1]

= n^{2}

Question 10.

Show that a_{1}, a_{2}, a_{3}, ………., a_{n} form an AP where a_{n},

is defined as below.

(i) a_{n} = 3 + 4n

(ii) a_{n} = 9 – 5n

Also find the sum of the first 15 terms in each case.

Solution:

(i) We have

a_{n} = 3 + 4n

Putting n = 1, 2, 3, 4 we get

a_{1} = 3 + 4 x 1 = 7

a_{2} = 3 + 4 x 2 = 11

a_{3} = 3 + 4 x 3 = 15

a_{4} = 3 + 4 x 4 = 19

…………………………………

…………………………………

…………………………………

a_{2} – a_{1} = 11 – 7 = 4

a_{3} – a_{2} = 15 – 11 = 4

a_{4} – a_{3} = 19 – 15 = 4

Heice the sequence is in AP with

a = 7, d = 4

Sum of first 15 terms

S_{15}= \(\frac { n }{ 2 } \) [2a + (n – 1)d]

S_{15} = \(\frac { 15 }{ 2 } \) [2 x 7 + (15 – 1) 4]

= \(\frac { 15 }{ 2 } \) [14 + 14 x 4]

= \(\frac { 15 }{ 2 } \) x [14 + 56]

= \(\frac { 15 }{ 2 } \) x 70 = 15 x 35 = 525

(ii) We have

a_{n} = 9 – 5n

Putting n = 1, 2, 3, 4 we get

a_{1} = 9 – 5 x 1 = 4

a_{2} = 9 – 5 x 2 = – 1

a_{3} = 9 – 5 x 3 = – 6

a_{4} = 9 – 5 x 4 = – 11

…………………………………

…………………………………

………………………………….

a_{2} – a_{1} = – 1 – 4 = 4

a_{3} – a_{2} = – 6 – ( – 1)

= – 6 + 1 = – 5

a_{4} – a_{3} = – 11 – (- 6)

= – 11 + 6 = – 5

Since the given sequence have same common differences hence it is in AP.

a = 4, d = – 5

Sum of first n term = \(\frac { n }{ 2 } \) [2a + (n – 1)d]

S_{15} = \(\frac { 15 }{ 2 } \) (2 x 4 + (15 – 1) x (- 5))

= \(\frac { 15 }{ 2 } \) [8 + 14 x (-5)]

= \(\frac { 15 }{ 2 } \) [8 – 70] = \(\frac { 15 }{ 2 } \) x – 62

= 15 x (-31) = – 465

Question 11.

If the sum of the first n terms of an AP is 4n – n^{2}, what is the first term (that is S_{1})? What is the

sum of first two terms? What is the second terms? Similarly, find the 3rd, the 10th and the nth terms.

Solution:

We have

Sum of the first n term = 4n – n^{2}

i.e. S_{n} = 4n – n^{2}

Putting n = 1, 2, 3, ……….

S_{1} = 4 x 1 – 1^{2} = 4 – 1 = 3

Therefore, a_{1} = S_{1} = 3

Hence the first term is 3

For n = 2

S_{2} = 4 x 2 x 2^{2} = 8 – 4 = 4

S2 is the sum of first two terms

Second term = S_{2} – S_{1}

a_{2} = 4 – 3 = 1

For, n = 3

S_{3} = 4 x 3 – 3^{2}

= 12 – 9 = 3

Therefore, third term = S_{3} – S_{2}

a_{3} = 3 – 4 = – 1

d = a_{2} – a_{1}

d = 1 – 3 = – 2

10th term = a_{10} = a + 9d

a_{10} = 3 + 9 x (-2)

= 3 – 2n + 2

= 5 – 2n

Question 12.

Find the sum of the first 40 positive integers divisible by 6.

Solution.

The first 40 integers divisible by 6 are 6, 12, 18, 24.

Here a_{2} – a_{1} = 126 = 6

a_{3} – a_{2} = 18 – 12 = 6

a_{4} – a_{3} = 24 – 18 = 6

Since difference of two consecutive terms is

same, so the list of integers divisible by fi is in AP.

Here a = 6; d = 6 and n = 40

Sum of first 40 positive integers = S_{40}

S_{40} = \(\frac { n }{ 2 } \) [2a + (n – 1)d]

S_{40} = \(\frac { 40 }{ 2 } \) [2 x 6 + (40 – 1) x 6]

= 20[12 + 39 x 6]

= 20[12 + 234]

= 20 x 248

= 4920

Question 13.

Find the sum of the first 15 multiples of 8.

Solution:

The first 15 multiples of 8 are

16, 24, 32, 40, …………

Here

a_{2} – a_{1} = 16 – 8 = 8

= 24 – 16 = 8

a_{3} – a_{2} = 32 – 24 = 8

Difference between the two consecutive terms

is same. Hence the List of multiples of 8 is in AP.

It is given

a = 8; d = 8 and n = 15

Sum of firs 15 multiples of 8 = S_{15}

S_{n} = \(\frac { n }{ 2 } \) [2a + (n – 1)d]

S_{15} = \(\frac { 15 }{ 2 } \) [2 x 8 + (15 – 1) x 8]

S_{15} = \(\frac { 15 }{ 2 } \) [16 + 14 x 8]

S_{15} = \(\frac { 15 }{ 2 } \) [16 + 112]

= \(\frac { 15 }{ 2 } \) [128]

= 15 x 64

= 960

Question 14.

Find the sum of odd numbers between O and 50.

Solution:

The odd numbers between 0 and 50 are 1, 3, 5, 7, ……………. 49

Here

a_{2} – a_{1} = 3 – 1 = 2

a_{3} – a_{2} = 5 – 3 = 2

a_{4} – a_{3} = 7 – 5 = 2

Difference between the two consecutive terms is same, Hence the list of odd numbers between 0 and 50 is in AP.

We have a = 1;d = 2 and l = 49

1 = a + (n – 1)d

49 = 1 + (n – 1) x 2

48 = (n – 1) x 2

n – 1 = \(\frac { 48 }{ 2 } \)

n – 1 = 24

n = 25

Sum of odd numbers between 0 to 50

S_{n} = \(\frac { n }{ 2 } \) [2a + (n – 1)d]

S_{25} = \(\frac { 25 }{ 2 } \) [2 x 1 + (25 – 1) x 2]

= \(\frac { 25 }{ 2 } \) x 50

= \(\frac { 25 }{ 2 } \) x 25

= 625

Question 15.

A contract an construction job specifies a penalty for delay of completion beyond a rertain date as follows: â‚¹ 200 for the first day, â‚¹ 250 for the second, â‚¹ 300 for the third day etc, the penalty for each succeeding day being â‚¹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he delays the work by 30 days?

Solution:

Penalty increases per day by â‚¹ 50.

We have 200, 250, 300, …………..

Here the list of penalty form an AP

a = 200.

d = 250 – 200 = 50

n = 30

Total money paid by contractor,

S_{n} = \(\frac { n }{ 2 } \) [2a + (n – 1)d]

S_{30} = \(\frac { 30 }{ 2 } \) [2 x 200 + (30 – 1) x 50]

= â‚¹ 15 [400 + 29 x 50]

= â‚¹ 15 x [400 + 1450]

= â‚¹ 15 x 1850

= â‚¹ 27750

Question 16.

A sum of â‚¹ 700 is to be used to gives even cash prizes to students of a school for their overall academic performance. If each prize is â‚¹ 20 less than its preceding prize, find the value of each of the prizes.

Solution:

Since value of each prize decreases by â‚¹ 20,

hence the value of seven succeeding cash prize form an AP.

It tint prize be â‚¹ x.

Then Second prize = x – 20

Third prize = x – 40

Fourth prize = x – 60

i.e. list of prize, x, x – 20, x – 40, z – 60, ……………

Here

a = x

d = x – 20 – x = – 20

n = 7

S_{7} = 700

We know that

S_{n} = \(\frac { n }{ 2 } \) [2a + (n – 1)d]

S_{7} = \(\frac { 7 }{ 2 } \) [2 x x + (7 – 1) x ( – 20)]

700 = \(\frac { 7 }{ 2 } \) [2x – 120]

700 = 7 [x – 60]

x – 60 = \(\frac { 100 }{ 7 } \)

x – 60 = 100

x = â‚¹ 160

Value of first prize = â‚¹ 160

Value of second prize

=x – 20 = 160 – 20

= â‚¹ 140

Value of third prize

= â‚¹ (140 – 20) = â‚¹ 120

Value of fourth prize

= â‚¹ (120 – 20) = â‚¹ 100

Value of fifth prize

= â‚¹ (100 – 20)= â‚¹ 80

Value of sixth prize

= â‚¹ (80 – 20) = â‚¹ 60

Value of seventh prize

= â‚¹ (60 – 20) = â‚¹ 40

Question 17.

In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g, a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class How many trees will be planted by the students?

Solution:

In class I there are three sections.

So number of trees planted by class 1 = 3.

Number of trees planted by class II

= 2 + 2 + 2 = 6

Number of trees planted by class III

= 3 + 3 + 3 = 9

Number of trees planted by class IV

= 4 + 4 + 4 = 12

In the same way,

Number of trees planted by class XII

= 12 + 12 + 12 = 36

We have 3, 6, 9, 12, …., 36

We observed that number of trees planted in

each class having same difference between two

consecutive classes. Therefore it forms an AP.

Here a = 3,d = 6 – 3 = 3 and n = 12

Total number of trees planted

S_{n} = \(\frac { n }{ 2 } \) [2a + (n – 1) d]

S_{12} = \(\frac { 12 }{ 2 } \) = [2 x 3 + (12 – 1) x 3]

= 6[6 + 33] = 6 x 39 = 234

Question 18.

A sapiral is made up ofÂ successive semicircles, with centres alternatively at A and B. starting with centres at A of radii 0.5 cm. 1.0cm 1.5 cm. 2.0 cm, … as shown in figure, What is the total length of such a spiral made up thirteen consecutive semicircles?

[Hint: Length of successive semicircles is l_{1}, l_{2}, l_{3}, l_{4} ……….. with centres A. B, A, B , ……….]

Solution:

Length of successive semicircles of radii, r_{1} – 0.5 cm, r_{2} = 1.0 cm, r_{3} = 1.5 cm, r_{4} = 2.0cm are l_{1}, = 0.5Ï€, l_{2} = 1 x Ï€, l_{3} = 1.5Ï€, l_{4} = 2Ï€, …

Let a_{1} = 0.5Ï€, a_{3} = 1.5Ï€, ………

a_{2} – a_{1} = Ï€ – 0,5Ï€ = 0.5Ï€ cm

a_{3} – a_{2} = 1.5Ï€ – Ï€ = 0.5Ï€ cm

a_{4} – a_{3} = 2Ï€ – 1.5cm = 0.5cm

Hence difference between two consecutive

terms is saine So the length of semicircles is in AP.

Here a = 0.5Ï€; d = 0.5Ï€ and Ï€ – 13

There fore total length

S_{n}= \(\frac { n }{ 2 } \) [2a + (n – 1)d]

S_{13} = \(\frac { 13 }{ 2 } \) [2 x 0.5Ï€ + (13 – 1) x 0.5Ï€] cm

= \(\frac { 13 }{ 2 } \) [Ï€ + 12 x 0.5Ï€] cm

= \(\frac { 13 }{ 2 } \) [Ï€ + 6Ï€] cm

= \(\frac { 13 }{ 2 } \) x 7Ï€ cm

= \(\frac { 13×7 }{ 2 } \) x \(\frac { 22 }{ 7 } \) cm

13 x 11 cm

= 143 cm

Question 19.

200 logs are stacked in the following manner: 20 logs in the bottom row. 19 in the next row, 18 in the row next Lo it and so on (see figure) in how many rows are the 200 logs placed and how many logs are in the top row?

Solution:

The number of logs in the bottom row, in the

next row, and in the next to it and so on are:

20, 19, 18, ………….

a_{2} – a_{1} = 19 – 20 = – 1

a_{3} – a_{2} = 18 – 19 = – 1

Hence two consecutive terms have same

difference So the above hat form an AP.

Here a = 20; d = – l and S_{n} = 200

We know that

S_{n}= \(\frac { n }{ 2 } \) [2a + (n – 1)d]

200 = \(\frac { n }{ 2 } \) [2 x 20 + (n – 1) x ( – 1)]

400 = n[40 – n + 1]

400 = n[41 – n]

400 = 41n – n^{2}

n_{2} – 41n +400 = 0

n^{2} – 25n – 16n + 400 = 0

n (n – 25) – 16 (n – 25) = 0

(n – 25) (n – 16) = 0

n = 25, or n = 16

Hence number of rows either 25 or 16.

Number of logs in nth row = a + (n – 1) d

Number of Logs in 25th row

= 20 + 24 x (-1)

= 20 – 24 = – 4

Which is not possible, therefore number of

rows i.e. n = 16

Number of Logaintop row = a_{16}

a_{n} = a + (n – 1)

a_{16} = a + (16 – 1)d

= 20 + 15 x (- 1)

= 20 – 15

= 5

Question 20.

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight unit. There are Len potatoes in the line (see Figure).

A competitor starts from the bucket, picks up the nearest potato, nina back with it, drop. it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in. and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run ?

[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 x 5 + 2 x (5 + 3)]

Solution:

To pick up the first potato, the distance run 2 x 5 m = 10 m

To pick up the second potato, the distance run = 2 x (5 + 3) = 16m

To pick up the third potato, the distance run = 2 (5 + 3 + 3) = 22 m and so on.

a_{2} – a_{1} = 16 – 10 = 6m

a_{3} – a_{2}= 22 – 16 – 6m

Hence difference between two consecutive terms are same. Therefore the list of distance run is in AP.

Here a = 10 m; d = 6m and n = 10

Therefore total distance run byu competitor = S_{10}

S_{n}= \(\frac { n }{ 2 } \) [2a + (n – 1)d]

S_{10} = \(\frac { 10 }{ 2 } \) [2 x 10 + (10 – 1) 6]

S_{10} = 5 [20 + 9 x 6]

S_{10} = 5 x [20 + 54]

S_{10} = 5 x 74

S_{10} = 370 m