NCERT Solutions Maths Class 10 Chapter 5

# NCERT Solutions Maths Class 10 Chapter 5

## NCERT Solutions Maths Class 10 Exercise 5.3

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) 1/15, 1/12, 1/10, …, to 11 terms.

Solution: (i) 2, 7, 12, …,  to 10 terms.

Here, first term (a) = 2, common difference (d) = 7 – 2 = 5 and n = 10

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

Sn = (10/2)[4 + (10 – 1)5]

= 5(4 + 45)

= 5 × 49

= 245

(ii) –37, –33, –29, …, to 12 terms.

Here, first term (a) = –37, common difference (d) = –33 – (–37) = 4 and n = 12

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

Sn = (12/2)[–74 + (12 – 1)4]

= 6(–74 + 44)

= 6 × (–30)

= –180

(iii) 0.6, 1.7, 2.8, …, to 100 terms.

Here, first term (a) = 0.6, common difference (d) = 1.7 – 0.6 = 1.1 and n = 100

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

Sn = (100/2)[1.2 + (100 – 1)1.1]

= 50(1.2 + 108.9)

= 50 × 110.1

= 5505

(iv) 1/15, 1/12, 1/10, …, to 11 terms.

Here, first term (a) = 1/15, common difference (d) = Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

2. Find the sums given below:

(i) 7 + 10½ + 14 + … + 84

(ii) 34 + 32 + 30 + … + 10

(iii) –5 + (–8) + (–11) + … + (–230)

Solution: (i) 7 + 10½ + 14 + … + 84

Here, first term (a) = 7, common difference (d) = Let the number of terms in the given AP be n.

Then, last term (l) = an = 84

So, we have to find n first.

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we have

[7 + (− 1) (3.5)] = 84

7 + 3.5− 3.5 = 84

3.5= 84 + 3.5 – 7

3.5= 80.5

= 23

Therefore, there are 23 terms in the given AP.

Applying formula, Sn = (n/2)[a + l] to find the sum of n terms of an AP, we have

S13 = (23/2)[7 + 84]

= (23/2)

= 1046.5

(ii) 34 + 32 + 30 + … + 10

Here, first term (a) = 34, common difference (d) = 32 – 34 = –2

Let the number of terms in the given AP be n.

Then, last term (l) = an = 10

So, we have to find n first.

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we have

[34 + (− 1) (−2)] = 10

34 – 2+ 2 = 10

−2= −26

= 13

Therefore, there are 13 terms in the given AP.

Applying formula, Sn = (n/2)[a + l] to find the sum of n terms of AP, we have

S13 = (13/2)[34 + 10]

= (13/2)

= 286

(iii) –5 + (–8) + (–11) + … + (–230)

Here, first term (a) = –5, common difference (d) = –8 – (–5) = –8 + 5 = –3

Let the number of terms in the given AP be n.

Then, last term (l) = an = –230

So, we have to find n first.

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we have

[−5 + (− 1) (−3)] = −230

−5 − 3+ 3 = −230

−3= −228

= 76

Therefore, there are 76 terms in the given AP.

Applying formula, Sn = (n/2)[a + l] to find the sum of n terms of AP, we have

S76 = (76/2)[−5 − 230]

= 38 (−235)

= −8930

3. In an AP:

(i) given = 5, = 3, an = 50, find n and Sn.

(ii) given = 7, a13 = 35, find d and S13.

(iii) given a12 = 37, = 3, find a and S12.

(iv) given a3 = 15, S10 = 125, find d and a10.

(v) given = 5, S9 = 75, find a and a9.

(vi) given = 2, = 8, Sn = 90, find n and an.

(vii) given = 8, an = 62, Sn = 210, find n and d.

(viii) given an = 4, = 2, Sn = −14, find n and a.

(ix) given = 3, = 8, = 192, find d.

(x) given = 28, = 144, and there are total 9 terms. Find a.

Solution: (i) given = 5, = 3, an = 50, find n and Sn.

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we have

an = 5 + (− 1)(3)

50 = 5 + 3− 3

48 = 3n

= 16

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S16 = (16/2)[10 + (16 – 1)3]

= 8(10 + 45)

= 8 × 55

= 440

Therefore, = 16 and Sn = 440

(ii) given = 7, a13 = 35, find d and S13.

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we have

a13= 7 + (13 − 1)d

35 = 7 + 12d

28 = 12d

= 7/3

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S13 = (13/2)[14 + (13 – 1)7/3]

= (13/2)(14 + 28)

= (13/2) × 42

= 273

Therefore, = 7/3 and S13 = 273

(iii) given a12 = 37, = 3, find a and S12.

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we have

a12+ (12 − 1)3

37 = + 33

= 4

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S12 = (12/2)[8 + (12 – 1)3]

= 6(8 + 33)

= 6 × 41

= 246

Therefore, = 4 and S12 = 246

(iv) given a3 = 15, S10 = 125, find d and a10.

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we have

a3 + (3 − 1) (d)

15 = + 2d

= 15 − 2d … (1)

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S10 = (10/2)[2a + (10 – 1)d]

125 = 5(2+ 9d)

25 = 2+ 9d … (2)

Putting the value of a from equation (1) in equation (2), we have

25 = 2(15 − 2d) + 9d = 30 − 4+ 9d

25 = 30 + 5d

–5 = 5d

= −1

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

a = 15 – 2(–1)

a = 15 + 2

a = 17

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we have

a10+ (10 − 1)d

Putting the values of a and d in the above equation, we get

a10 = 17 + 9 × (–1)

= 17 – 9

= 8

Therefore, = −1 and a10 = 8

(v) given = 5, S9 = 75, find a and a9.

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S9 = (9/2)[2a + (9 – 1)5]

75 = (9/2)[2a + 40]

150 = 18+ 360

−210 = 18a

= −35/3

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we have

a9 = −35/3 + (9 − 1) (5)

= −35/3 + 40

= 85/3

Therefore, = −35/3 and a9 = 85/3

(vi) given = 2, = 8, Sn = 90, find n and an.

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

90 = (n/2)[4 + (n – 1)8]

90 = (n/2)[4 + 8n – 8]

90 = (n/2)[8n – 4]

8n2 – 4n – 180 = 0

2n2n – 45 = 0

2n2 – 10n + 9n – 45 = 0

2n(− 5) + 9(− 5) = 0

(− 5) (2+ 9) = 0

= 5, −9/2

[We discard the negative value of n because n cannot be negative or a fraction.]

Therefore, = 5

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we have

a5 = 2 + (5 − 1) (8) = 2 + 32 = 34

Therefore, = 5 and an = 34.

(vii) given = 8, an = 62, Sn = 210, find n and d.

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we have

62 = 8 + (− 1) (d)

62 = 8 + nd − d

nd – = 54

nd = 54 + d … (1)

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

210 = (n/2)[16 + (n – 1)d]

210 = (n/2)[16 + ndd]

Putting the value of nd from equation (1) in the above equation, we have

210 = (n/2)[16 + 54 + dd]

210 = (n/2)

210 = n × 35

= 6

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

6= 54 +

5= 54

= 54/5

Therefore, = 6 and = 54/5.

(viii) given an = 4, = 2, Sn = −14, find n and a.

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we have

4 = + (− 1) (2)

4 = + 2– 2

6 = + 2n

= 6 − 2n … (1)

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

−14 = (n/2)[2a + (n – 1)2]

−14 = (n/2)[2a + 2n – 2]

−28 = n[2a + 2n – 2]

Putting the value of a from equation (1) in the above equation, we get

−28 = n[2(6 − 2n) + 2– 2]

−28 = n(12 − 4+ 2− 2)

−28 = n(10 − 2n)

2n2 – 10n – 28 = 0

n2 – 5n – 14 = 0

n2 – 7n + 2n – 14 = 0

n(− 7) + 2(− 7) = 0

(+ 2) (− 7) = 0

= −2, 7

[We discard the negative value of n because it cannot be negative.]

Therefore, = 7.

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

= 6 − 2= 6 – 2(7) = 6 – 14 = −8

Therefore, = 7 and = −8.

(ix) given = 3, = 8, = 192, find d.

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

192 = (8/2)[6 + (8 − 1)d]

192 = 4(6 + 7d)

192 = 24 + 28d

168 = 28

= 6

(x) given = 28, = 144, and there are total 9 terms. Find a.

Applying formula, Sn = (n/2)[a + l] to find the sum of n terms of an AP, we get

144 = (9/2)[+ 28]

288 = 9[+ 28]

32 = + 28

= 4

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

Solution: Here, first term (a) = 9, common difference (d) = 17 – 9 = 8 and Sn = 636

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

636 = (n/2)[18 + (− 1) (8)]

1272 = n(18 + 8− 8)

1272 = n(10 + 8n)

1272 = 10n + 8n2

8n2 + 10n – 1272 = 0

4n2 + 5n – 636 = 0

Comparing equation 4n2 + 5n – 636 = 0 with the general form an2 + bn + c = 0, we get

= 4, = 5 and = −636

Applying quadratic formula, and putting the values of a, b and c, we get

[Here, we discard negative value of n because n cannot be negative.]

Therefore, = 12

Hence, 12 terms of the given sequence give a sum of 636.

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: Here, first term (a) = 5, last term (l) = 45 and Sn = 400

Applying formula, Sn = (n/2)[a + l] to find the sum of n terms of an AP, we get

400 = (n/2)[5 + 45]

400 = (n/2) × 50

400 = n × 25

n = 16

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

400 = (16/2)[10 + (16 – 1)d]

400 = 8(10 + 15d)

400 = 80 + 120d

320 = 120d

= 320/120

= 8/3

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, first term (a) = 17, last term (l) = 350 and common difference (d) = 9

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we get

350 = 17 + (− 1)(9)

350 = 17 + 9− 9

342 = 9

= 38

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S38 = (38/2)[34 + (38 – 1)9]

S38 = 19(34 + 333) = 6973

Therefore, there are 38 terms and their sum is equal to 6973.

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

Solution:  It is given that 22nd term is 149

It means a22 = 149

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we get

149 = + (22 − 1)(7)

149 = + 147

= 2

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S22= (22/2)[4 + (22 – 1)7)

S22= 11(4 + 147)

S22= 1661

Therefore, the sum of first 22 terms of the AP is 1661.

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

Solution: It is given that the second and third terms of an AP are 14 and 18 respectively.

It means that a2 = 14 and a3 = 18

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we get

14 = + (2 − 1)d

14 = d … (1)

And, 18 = + (3 − 1)d

18 = + 2d … (2)

These are the linear equations in two variables.

Using equation (1), we get = 14 − d

Putting the value of a in equation (2), we get

18 = 14 – + 2d

= 4

Therefore, common difference (d) = 4

Putting the value of d in equation (2), we get

18 = + 2(4)

= 10

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S51 = (51/2)[20 + (51 – 1)4]

S51 = (51/2)[20 + 200]

S51 = (51/2)

S51 = 51 × 110

S51 = 5610

Therefore, sum of first 51 terms of an AP is equal to 5610.

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: It is given that the sum of first 7 terms of an AP is 49 and the sum of first 17 terms is 289.

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

49 = (7/2)[2a + (7 – 1)d]

98 = 7(2+ 6d)

14 = 2+ 6d

7 = + 3

= 7 − 3d … (1)

And, 289 = (17/2)[2a + (17 – 1)d]

578 = 17(2+ 16d)

34 = 2+ 16d

17 = + 8d … (2)

Putting the value of a from equation (1) in equation (2), we get

17 = 7 − 3+ 8d

10 = 5

= 2

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

= 7 − 3

= 7 – 3(2)

= 7 – 6

= 1

Again applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

Sn = (n/2)[2(1) + (n – 1)2]

Sn = (n/2)[2 + 2n – 2]

Sn = (n/2) × 2n

Sn n2

Therefore, the sum of n terms of the AP is n2.

10. Show that a1, a2, …, an … form an AP where an is defined as below:

(i) an = 3 + 4n

(ii) an = 9 – 5n

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

Solution:  (i) We have to show that a1, a2 …, an form an AP where an = 3 + 4n

Let us calculate the values of a1, a2 …, using an = 3 + 4n

a1 = 3 + 4(1) = 3 + 4 = 7

a2 = 3 + 4(2) = 3 + 8 = 11

a3 = 3 + 4(3) = 3 + 12 = 15

a4 = 3 + 4(4) = 3 + 16 = 19

So, the sequence is of the form 7, 11, 15, 19, …

Let us find the difference between two consecutive terms of the above sequence.

11 – 7 = 4, 15 – 11 = 4, 19 – 15 = 4

Since the difference between two consecutive terms is the same which means that the terms a1a2 …, an form an AP.

We have the sequence: 7, 11, 15, 19 …

Here, first term (a) = 7 and common difference (d) = 4

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S15 = (15/2)[14 + (15 – 1)4]

S15 = (15/2)[14 + 56]

S15 = (15/2) × 70

S15 = 15 × 35

S15 = 525

Therefore, the sum of the first 15 terms of the AP is 525.

(ii) We have to show that a1, a2 …, an form an AP where an = 9 – 5n

Let us calculate the values of a1, a2 …, an using an = 9 – 5n

a1 = 9 – 5(1) = 9 – 5 = 4

a2 = 9 – 5(2) = 9 – 10 = −1

a3 = 9 – 5(3) = 9 – 15 = −6

a4 = 9 – 5(4) = 9 – 20 = −11

So, the sequence is of the form 4, −1, −6, −11, …

Let us find the difference between two consecutive terms of the above sequence.

–1 – (4) = –5

–6 – (–1) = –6 + 1 = –5

–11 – (–6) = –11 + 6 = –5

Since the difference between two consecutive terms is the same which means that the terms a1a2 …, an form an AP.

We have the sequence: 4, −1, −6, −11, …

Here, first term (a) = 4 and common difference (d) = −5

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S15 = (15/2)[8 + (15 – 1)(–5)]

S15 = (15/2)[8 – 70]

S15 = (15/2)[–62]

S15 = 15 × (–31)

S15 = –465

Therefore, the sum of the first 15 terms of the AP is –465.

11. If the sum of the first n terms of an AP is (4n n2), what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Solution: It is given that the sum of the first n terms of an AP is (4n n2).

It means Sn = 4nn2

Let us calculate S1 and S2 using Sn = 4nn2

S1 = 4(1) – (1)2 = 4 – 1 = 3

S2 = 4(2) – (2)2 = 8 – 4 = 4

First term (a) = S1= 3 … (1)

Now, let us find the common difference.

We can write any AP in the form of general terms like a, a + d, a + 2d

We have calculated that S2 = 4

Therefore, we can say that a + (a + d) = 4

2a + d = 4

Putting the value of a from equation (1), we get

2(3) + d = 4

6 + d = 4

d = –2

Using formula, an = a + (n – 1)d, to find the nth term of an AP, we get

a2 = a + (2 – 1)d

= 3 + (2 – 1)(–2)

= 3 – 2 = 1

a3 = a + (3 – 1)d

= 3 + (3 – 1)(–2)

= 3 – 4 = –1

a10 = a + (10 – 1)d

= 3 + (10 – 1) (–2)

= 3 – 18 = –15

an = a + (n – 1)d

= 3 + (n – 1)(–2)

= 3 – 2n + 2

= 5 – 2n

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

Solution:  The first 40 positive integers divisible by 6 are 6, 12, 18, 24 … up to 40 terms.

Therefore, we have to find the sum of 40 terms of the above sequence.

Here, first term (a) = 6, common difference (d) = 12 – 6 = 6 and n = 40

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S40 = (40/2)[12 + (40 – 1)6]

= 20(12 + 39 × 6)

= 20(12 + 234)

= 20 × 246 = 4920

13. Find the sum of the first 15 multiples of 8.

Solution: The first 15 multiples of 8 are 8, 16, 24, 32 … up to 15 terms.

Therefore, we have to find the sum of 15 terms of the above sequence.

Here, first term (a) = 8, common difference (d) = 16 – 8 = 8 and n = 15.

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S15 = (15/2)[16 + (15 – 1)8]

S15 = (15/2)[16 + 14 × 8]

S15 = (15/2)[16 + 112]

S15 = (15/2)128

S15 = 15 × 64

S15 = 960

14. Find the sum of the odd numbers between 0 and 50.

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

It is an AP because the difference between two consecutive terms is the same.

Here, first term (a) = 1, common difference (d) = 3 – 1 = 2 and last term (l) = 49

We do not know that how many odd numbers are there between 0 and 50.

Therefore, we have to find n first.

Using formula, an = a + (n − 1) d, to find the nth term of an AP, we get

49 = 1 + (n − 1)2

49 − 1 = 2n – 2

48 = 2n − 2

2n = 50

n = 25

Applying formula, Sn = (n/2)[a + l] to find the sum of n terms of an AP, we get

S25 = (25/2)[1 + 49]

S25 = (25/2)

S25 = 25 × 25

S25 = 625

Therefore, the sum of the odd numbers between 0 and 50 is 625.

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

Solution: Penalty for the first day = Rs 200, penalty for the second day = Rs 250 and

penalty for the third day = Rs 300

It is given that the penalty for each succeeding day is Rs 50 more than the preceding day.

It forms an arithmetic progression because the difference between any two consecutive terms is the same.

We have to find how much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

So, we have an AP of the form 200, 250, 300, 350 … up to 30 terms.

Here, first term (a) = 200, common difference (d) = 50 and n = 30

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

Sn = (30/2)(400 + (30 – 1)50)

Sn = 15(400 + 29 × 50)

Sn = 15(400 + 1450) = 27750

Therefore, the penalty for 30 days is Rs 27,750.

16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

Solution: It is given that the sum of seven cash prizes is equal to Rs 700.

And each prize is Rs 20 less than its preceding prize.

Let the value of first prize be Rs a.

Then the value of second prize = Rs (a − 20)

The value of third prize = Rs (a − 40)

So, we have the sequence of the form:

a, a − 20, a − 40, a – 60 …

It is an arithmetic progression because the difference between two consecutive terms is the same.

Here, first term = a, common difference (d) = (a – 20) – a = –20

n = 7 (Because there are total of seven prizes)

It is given that S7 = Rs 700

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S7 = (7/2)[2a + (7 – 1)(–20)]

700 = (7/2)[2a – 120]

200 = 2a – 120

320 = 2a

a = 160

Therefore, the value of first prize = Rs 160

The value of second prize = 160 – 20 = Rs 140

The value of third prize = 140 – 20 = Rs 120

The value of fourth prize = 120 – 20 = Rs 100

The value of fifth prize = 100 – 20 = Rs 80

The value of sixth prize = 80 – 20 = Rs 60

The value of seventh prize = 60 – 20 = Rs 40

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 two trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution: It is given that the number of trees planted by each class is equal to class number. There are three sections of each class.

The number of trees planted by class I = number of sections × 1 = 3 × 1 = 3

The number of trees planted by class II = number of sections × 2 = 3 × 2 = 6

The number of trees planted by class III = number of sections × 3 = 3 × 3 = 9

Therefore, we have sequence of the form 3, 6, 9 … up to 12 terms.

To find the total number of trees planted by all the students, we have to find the sum of the sequence 3, 6, 9, 12 … up to 12 terms.

Here, first term (a) = 3, common difference (d) = 6 – 3 = 3 and n = 12

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S12 = (12/2)[6 + (12 – 1)3]

S12 = 6(6 + 33)

S12 = 6 × 39

S12 = 234

Hence, all the students planted 234 trees.

18. A spiral is made up of successive semicircles, with centres alternatively at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … What is the total length of such a spiral made up of thirteen consecutive semicircles.

Length of semicircle of radii 0.5 cm = π(0.5) cm

Length of semicircle of radii 1.0 cm = π(1.0) cm

Length of semicircle of radii 1.5 cm = π(1.5) cm

Therefore, we have the sequence of the form:

π(0.5), π(1.0), π(1.5) … up to 13 terms [Since there are total thirteen semicircles.]

To find the total length of the spiral, we have to find the sum of the sequence π(0.5), π(1.0), π (1.5) … up to 13 terms.

Total length of the spiral = π(0.5) + π(1.0) + π(1.5) … up to 13 terms

Total length of the spiral = π(0.5 + 1.0 + 1.5) … up to 13 terms … (1)

The sequence 0.5, 1.0, 1.5 … up to 13 terms is an arithmetic progression.

Let us find the sum of this sequence.

Here, first term (a) = 0.5, common difference (d) = 1.0 – 0.5 = 0.5 and n = 13

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S13 = (13/2)[1 + (13 – 1)0.5]

S13 = 6.5[1 + 6]

S13 = 6.5 × 7

S13 = 45.5

Therefore, 0.5 + 1.0 + 1.5 + 2.0 … up to 13 terms = 45.5

Putting this value in equation (1), we get

Total length of the spiral = π(0.5 + 1.5 + 2.0 + … up to 13 terms)

π(45.5) = (22/7)(45.5) = 143 cm

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 to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Solution: The total number of logs in the bottom row = 20

The number of logs in the next row = 19

The number of logs in the next to next row = 18

Therefore, we have the sequence of the form 20, 19, 18 …

Here, first term (a) = 20, common difference (d) = 19 – 20 = –1

We have to find that how many rows make total of 200 logs. Let there are a total of n rows.

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

200 = (n/2)[40 + (n – 1) (–1)]

400 = n(40 – n + 1)

400 = 40nn2 + n

n2 – 41n + 400 = 0

We can factorize it to solve the equation.

n2 − 25n − 16n + 400 = 0

n(n − 25) – 16(n − 25) = 0

(n − 25) (n − 16)

n = 25, 16

We discard n = 25 because we cannot have more than 20 rows in the sequence.

The sequence is of the form: 20, 19, 18 …

Here, n = 16 which means 16 rows make a total number of logs equal to 200.

We also have to find the number of logs in the 16th row.

Applying formula, Sn = (n/2)[a + l] to find the sum of n terms of an AP, we get

200 = 8(20 + l)

200 = 160 + 8I

40 = 8

= 5

Therefore, there are 5 logs in the topmost row and there are a total of 16 rows.

20. In a potato race, a bucket is placed at the starting point, which is 5 metres from the first potato, and the other potatoes are placed 3 metres apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops 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?

Solution: The distance of first potato from the starting point = 5 metres

Therefore, the distance covered by competitor to pick up first potato and put it in bucket = 5 × 2 = 10 metres

The distance of second potato from the starting point = 5 + 3 = 8 metres

[All the potatoes are 3 metres apart from each other.]

Therefore, the distance covered by competitor to pick up 2nd potato and put it in bucket = 8 × 2= 16 metres

The distance of third potato from the starting point = 8 + 3 = 11 metres

Therefore, the distance covered by competitor to pick up 3rd potato and put it in bucket = 11 × 2 = 22 metres

Therefore, we have a sequence of the form: 10, 16, 22 … up to 10 terms.

[There are a total of ten terms because there are ten potatoes.]

To calculate the total distance covered by the competitor, we have to find:

10 + 16 + 22 + … up to 10 terms.

Here, first term (a) = 10, common difference (d) = 16 – 10 = 6 and n = 10

Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have

S10 = (10/2)[20 + (10 – 1)6]

S10 = 5(20 + 54)

S10 = 5 × 74

S10 = 370

Therefore, the total distance covered by the competitor is 370 metres.