Haryana State Board HBSE 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 Textbook Exercise Questions and Answers.

## Haryana Board 10th Class Maths Solutions 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) \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots\) to 11 terms.

Solution :

(i) The given sequence of AP is : 2, 7, 12, …………….., to 10 terms

Here, a = 2

d = a_{2} – a_{1}

=7 – 2 = 5 and n = 10

We know that

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

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

⇒ S_{10} = 5[4 + 9 × 5]

⇒ S_{10} = 5 [4 + 45]

⇒ S_{10} = 5 × 49 = 245

Hence, the sum of 10 terms of given AP = 245.

(ii) The given sequence of AP is : – 37, – 33, – 29, …………., to 12 terms

Here, a = – 37

d = a_{2} – a_{1}

= – 33 – (- 37)

= – 33 + 37 = 4

and n = 12

We know that

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

⇒ S_{12} = \(\frac{n}{2}\) [2 × (- 37) + (12 – 1) × 4]

⇒ S_{12} = 6[- 74 + 11 × 4]

⇒ S_{12} = 6[- 74 + 44]

⇒ S_{12} = 6 × (- 30) = – 180

Hence, the sum of 12 terms of given AP = – 180.

(iii) The given sequence of AP is : 0.6, 1.7, 2.8, …………., to 100 terms.

Here, a = 0.6

d = a_{2} – a_{1}

= 1.7 – 0.6 = 1.1 and n = 100

We know that

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

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

⇒ S_{100} = 50 [1.2 + 99 × 1.1]

⇒ S_{100} = 50[1.2 + 108.9]

⇒ S_{100} = 50 × 110.1 = 5505

Hence, the sum of 100 terms of given AP = 5505.

(iv) The given sequence of AP is : \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots\) to 11 terms

Hence, the sum of 11 terms of given AP = \(\frac{33}{20}\).

Question 2.

Find the sum of the following APs given below :

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

(ii) 34 + 32 + 30 + …………. + 10

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

Solution:

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

It is an AP.

Here, a = 7

d = a_{2} – a_{1}

d = 10\(\frac{1}{2}\) – 1

d = \(\frac{21-14}{2}\) = \(\frac{7}{2}\)

and l = a_{n} = 84

⇒ a + (n – 1)d = 84

⇒ 7 + (n – 1) × \(\frac{7}{2}\) = 84

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

\(\frac{7}{2}\) n – \(\frac{7}{2}\) = 77

\(\frac{7}{2}\) n = 77 + \(\frac{7}{2}\)

\(\frac{7}{2}\) n = \(\frac{154+77}{2}\)

⇒ n = \(\frac{161}{7}\) = 23

We know that

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

S_{n} = \(\frac{23}{2}\) [7 + 84]

S_{23} = \(\frac{23}{2}\) × 91

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

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

(ii) We have, 34 + 32 + 30 + …………….. + 10.

It is an AP.

Here, a = 34

d = a_{2} – a_{1}

= 32 – 34 = – 2 and

l = a_{n} = 10

⇒ a + ( n – 1 )d = 10

⇒ 34 + (n – 1) × (- 2) = 10

⇒ – 2n + 2 = 10 – 34

⇒ – 2n + 2 = – 24

⇒ – 2n = – 24 – 2

⇒ – 2n = – 26

⇒ n = \(\frac{-26}{-2}\) = 13

We know that

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

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

= \(\frac{13}{2}\) × 44

S_{13} = 286

Hence S_{13} = 286.

(ill) Wehave, – 5 + (- 8) + (- 11) + …………… + (- 230)

It is an AP.

Here, a = – 5

d = a_{2} – a_{1}

= (- 8) – (- 5)

and l = a_{n} = – 230

⇒ a + (n – 1) d = – 230

⇒ – 5 + (n – 1) × (- 3) = – 230

⇒ – 3n + 3 = – 230 + 5

⇒ – 3n + 3 = – 225

⇒ – 3n = – 225 – 3

⇒ – 3n = – 228

⇒ n = \(\frac{-228}{-3}\) = 76

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

S_{76} = \(\frac{76}{2}\) [- 5 – 230] = 38 × (- 235)

S_{76} = – 8930

Hence, S_{76} = – 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 a_{12} = 37, d = 3, find a and S_{12}.

(iv) given a_{3} = 15, S_{10} = 125, find d and a_{10}.

(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_{n} = 192, find d.

(x) given l = 28, S_{n} = 144, and there are total 9 terms. Find a.

Solution :

(i) We have, a = 5, d = 3, a_{n} = 50

We know that

a + (n – l)d = a_{n}

⇒ 5 + (n – 1) × 3 = 50

⇒ 5 + 3n – 3 = 50

⇒ 2 + 3n – 50

⇒ 3n = 50 – 2 = 48

⇒ n = \(\frac{48}{3}\) = 16

We know that

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

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

S_{16} = 8 [10 + 15 × 3]

= 8 [10 + 45]

S_{16} = 8 × 55 = 440

Hence, n = 16, S_{16} = 440.

(ii) We have, a = 7, a_{13} = 35

We know that

a + (n – 1) d = a_{n}

7 + (13 – 1) d = 35

7 + 12d = 35

12d = 35 – 7 = 28

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

We know that

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

[ l = a_{13} = 35]

S_{13} = \(\frac{13}{2}\) [7 + 35]

= \(\frac{13}{2}\) × 42 = 273

Hence d = \(\frac{7}{3}\), S_{13} = 273.

(iii) We have, d = 3, a_{12} = 37

We know that

a + (n – 1 )d = a_{n}

⇒ o + (12 – 1) × 3 = 37

⇒ a + 11 × 3 = 37

⇒ a + 33 = 37

⇒ a = 37 – 33 = 4

We know that

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

S_{12} = \(\frac{12}{2}\) [4 + 37] = 6 × 41

[∵ l = a_{12} = 37]

S_{12} = 246.

Hence, a = 4, a_{12} = 246.

(iv) We have, a_{3} = 15, S_{10} = 126

Now, a_{3} – 15

a + (3 – 1) d = 15

[∵ a_{n} = a + (n – 1)d]

a + 2d = 15 ……………(1)

and S_{10} = 125

\(\frac{n}{2}\) [2a + (10- 1)d] = 125

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

5 [2a + 9d] = 125

2a + 9d = \(\frac{125}{5}\)

2a + 9d = 25 ………………(2)

Multiplying equation (1) by 2 and subtracting (2) from it, we get

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

a + 2 × (- 1) = 16

a – 2 = 15

a = 15 + 2

a = 17

Now, a_{10} = 17 + (10 – 1) × (- 1)

a_{10} = 17 + 9 × (- 1)

a_{10} = 17 – 9

a_{10} = 8

Hence, d = – 1 and a_{10} = 8.

(v) We have, d = 5, S_{9} = 75

⇒ \(\frac{9}{2}\) [2a + (9 – 1) × 5] = 75

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

⇒ \(\frac{9}{2}\) [2a + 8 × 5] = 75

(vi) We have, a = 2, d = 8, S_{n} = 90

⇒ \(\frac{n}{2}\) [2a + (n – 1) d] = 90

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

\(\frac{n}{2}\) [2 × 2 + (n – 1)d] = 90

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

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

⇒ \(\frac{n}{2}\) [4 + 8n – 8] = 90

⇒ \(\frac{n}{2}\) [8n – 4] = 90 × 2

⇒ 8n^{2} – 4n = 180

⇒ 2n^{2} – n = 45

⇒ 2n^{2} – n – 45 = 0

⇒ 2n^{2} – (10 – 9)n – 45 = 0

⇒ 2n^{2} – 10n + 9n – 45 =0

⇒ 2n(n – 5) + 9(n – 5) = 0

[∵ 2 × – 45 = – 90

10 × – 9 = – 90

10 – 9 = 1]

⇒ (n – 5) (2n + 9) = 0

n – 5 = 0 or 2n + 9 = 0

n = 5 or n = – \(\frac{9}{2}\)

Reject n = – \(\frac{9}{2}\) [∵ n ≠ – \(\frac{9}{2}\)]

So, n = 5

Now, a_{n} = a + (n – 1)d

⇒ a_{5} = 2 + (5 – 1) × 8

⇒ a_{5} = 2 + 4 × 8

⇒ a_{5} = 2 + 32

⇒ a_{5} = 34

Hence, n = 5 and a_{5} = 34.

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

Now, S_{n} = 210

\(\frac{9}{2}\) [a + l] = 210

\(\frac{9}{2}\) [8 + 62] = 210 [∵ l = a_{n} = 62]

\(\frac{n}{2}\) × 70 = 210

35n = 210

n = \(\frac{210}{35}\) = 6

Now, a_{n} = 62

⇒ 8 + (6 – 1) d = 62

⇒ [a_{n} = a + (n – 1)d]

⇒ 8 + 5d = 62

⇒ 5d = 62 – 8

⇒ 5d = 54

⇒ d = – \(\frac{54}{5}\)

Hence, n = 6, d = \(\frac{54}{5}\).

(viii) We have, d = 2, S_{n} = – 14, a_{n} = 4

⇒ a + (n – 1) × 2 = 4[∵ a_{n} = a + (n – 1) d]

⇒ a + 2n – 2 = 4

⇒ a + 2n = 4 + 2

⇒ a + 2n = 6

⇒ a = 6 – 2n ……………….(i)

Now, S_{n} = – 14

⇒ \(\frac{n}{2}\) [a + l] = – 14

⇒ n[6 – 2n + 4] = – 14 × 2

[From (1) Put a = 6 – 2n]

⇒ n [10 – 2n] = – 28 [∵ l = a_{n} = 4]

⇒ 10n – 2n^{2} = – 28

⇒ 2n^{2} – 10n – 28 = 0

⇒ n^{2} – 5n – 14 = 0

⇒ n^{2} – (7 – 2)n – 14 = 0

⇒ n^{2} – 7n + 2n – 14 = 0

[∵ 1 × – 14 = – 14

7 × – 2 = – 14

7 – 2 = 5]

⇒ n(n – 7) + 2(n – 7) = 0

⇒ (n – 7)(n + 2) = 0

⇒ n – 7 = 0 or n + 2 = 0

⇒ n = 7 or n = – 2

Reject n = – 2 [∵ n ≠ – 2]

So, n = 7

Substituting the value of n in equation (1), we

a = 6 – 2 × 7

⇒ a = 6 – 14

⇒ a = – 8.

Hence, n = 7, a = – 8.

(ix) We have, a = 3, n = 8, S_{n} = 192

i.e. S_{8} = 192

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

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

⇒ 4 [6 + 7d] = 192

⇒ 6 + 7d = 48

⇒ 7d = 48 – 6

⇒ 7d = 42

⇒ d = \(\frac{42}{7}\) = 6

Hence, d = 6.

(x) We have, n = 9, l = 28, S_{n} = 144

\(\frac{n}{2}\) [a + l] = 144

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

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

a + 28 = 32

a = 32 – 28 = 4

Hence, a = 4.

Question 4.

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

Solution:

The given sequence of AP is: 9, 17, 25, ………….

Here, a = 9

d = a_{2} – a_{1}

= 17 – 9 = 8

and S_{n} = 636 (Given)

\(\frac{n}{2}\) [2a + (n – 1)d] = 636

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

n [18 + 8n – 8] = 636 × 2

n [10 + 8n] = 1272

10n + 8n^{2} = 1272

8n^{2} + 10n – 1272 = 0

4n^{2} + 5n – 636 = 0

4n^{2} + (53 – 48)n – 636 = 0

[∵ – 636 × 4 = – 2544

53 × – 48 = – 2544

53 – 48 = 5]

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 = – \(\frac{53}{4}\) or n = 12

Reject, n = – \(\frac{53}{4}\)

[∵ n cannot be fraction]

So, n = 12.

Hence, 12 terms of the given AP must be taken to give a sum of 636.

Question 5.

The first term of an AP is 5, the last term 1845 and the sum is 400. Find the number of terms and the common difference.

Solution :

We have, a = 5, l = a_{n} = 45, S_{n} = 400

\(\frac{n}{2}\) [a + l] = 400

\(\frac{n}{2}\) [15 + 45] = 400

n × 50 = 400 × 2

50n = 800

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

n = 16

Now, a_{n} = 45

a + (n – 1) d = 45

5 + (16 – 1)d = 45

5 + 15d = 45

15d = 45 – 5

15d = 40

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

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

Hence, n = 16, d = \(\frac{8}{3}\).

Question 6.

The first and last terms of an AP are 17 and 350 respectively. 1f the common difference is 9, how many terms are there and what is their sum?

Solution:

We have, a = 17, d = 9, l = a_{n} = 350

a + (n – 1)d = 350

17 + (n – 1)9 = 350

17 + 9n – 9 = 350

8 + 9n = 350

9n = 350 -8

9n = 342

n = \(\frac{342}{9}\) = 38

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

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

S_{38} = 19 × 367

S_{38} = 6973

Hence, the number of terms = 38

and their sum = 6973.

Question 7.

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

Solution :

We have, d = 7, n = 22, a_{22} = 149

a + (22 – 1) × 7 = 149

[∵ a_{n} = a + (n – 1)d]

a + 21 × 7 = 149

a + 147 = 149

a = 149 – 147

a = 2

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

S_{22} = \(\frac{22}{2}\) [2 + 149]

[∵ l = a_{n} = 149]

S_{22} = 11 × 151

S_{22} = 1661

Hence, S_{22} = 1661.

Question 8.

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

Solution :

We have, n = 51, a_{2} = 14, a_{3} = 18

d = a_{2} – a_{1}

⇒ d = 18 – 14

d = 4

Therefore, a_{3} = a – d

= 14 – 4 = 10

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

S_{51} = \(\frac{51}{2}\) [2 × 10 + (51 – 1) × 4]

S_{51} = \(\frac{51}{2}\) [20 + 50 × 4]

⇒ S_{51} = \(\frac{51}{2}\) [20 + 200]

⇒ S_{51} = \(\frac{51}{2}\) [220]

⇒ S_{51} = 51 × 110

⇒ S_{51} = 5610

Hence, S_{51} = 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 :

We have, S_{7} = 49

⇒ \(\frac{7}{2}\) [2a + (7 – 1)5] = 49

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

2a + 6d = \(\frac{49 \times 2}{7}\)

2a + 6d = 14

a + 35 = 7 ……………..(1)

S_{17} = 289

\(\frac{17}{2}\) [2a + (17 – 1) 5] = 289

2a + 16d = \(\frac{289 \times 2}{17}\)

2a + 16d = 34

a + 8d = 17

Subtracting equation (2) from equation (1), we get

Putting the value of 5 in equation (1), we get

a + 3 × 2 = 7

⇒ a + 6 = 7

⇒ a = 7 – 6

⇒ a = 1

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

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

⇒ S_{n} = \(\frac{n}{2}\) [2 + 2n – 2]

⇒ S_{n} = \(\frac{n}{2}\) [2n]

⇒ S_{n} = n^{2}

Hence S_{n} = n^{2}.

Question 10.

Show that a_{1}, a_{2}, …………., 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,…………. in this equation, we get

a_{1} = 3 + 4 × 1

⇒ a_{1} = 7

a_{2} = 3 + 4 × 2

⇒ a_{2} = 3 + 8

⇒ a = 11

a_{3} = 3 + 4 × 3

⇒ a_{3} = 3 + 12

a_{3} = 15

a_{4} = 3 + 4 × 4

⇒ a_{4} = 3 + 16

⇒ a_{4} = 19 …………… and so on.

Therefore, the sequence is : 7, 11, 15, 19, …………..

a_{2} – a_{1} = 11 – 7 = 4

a_{3} – a_{2} = 15 – 11 = 4

a_{4} – a_{3} = 19 – 15 = 4

∵ a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3}

The given sequence forms an AP Proved.

Here a = 7, d = a_{2} – a_{1} = 11 – 7 = 4, n = 15

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

⇒ S_{15} = \(\frac{15}{2}\) [2 × 7 + (15 – 1)

⇒ S_{15} = \(\frac{15}{2}\) [14 + 14 × 4]

⇒ S_{15} = \(\frac{15}{2}\) [14 + 56]

⇒ S_{15} = \(\frac{15}{2}\) × 70

S_{15} = 15 × 35

S_{15} = 525

Hence, S_{15} = 525.

(ii) We have, a_{n} = 9 – 5n

Putting n = 1, 2, 3, 4, …………. in this equation,

a_{1} = 9 – 5 × 1

a_{1} = 9 – 5

⇒ a_{1} = 4

a_{2} = 9 – 5 × 2

⇒ a_{2} = 9 – 10

⇒ a_{2} = -1

a_{3} = 9 – 5 × 3

⇒ a_{3} = 9 – 15

a_{3} = – 6

a_{4} = 9 – 5 × 4

⇒ a_{4} = 9 – 20

⇒ a_{4} = – 11

Therefore, the sequence is : 4, – 1, – 6, – 11,…………..

a_{2} – a_{1} = -1 – 4 = – 5

a_{3} – a_{2} = – 6 – (- 1)

= – 6 + 1 = – 5

a_{4} – a_{3} = – 11 – (- 6)

= – 11 + 6 = – 5

∵ a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3}

∴ The given sequence forms an AP. Proved

Here, a = 4, d = a_{2} – a_{2} = – 1 – 4 = – 5, n = 15

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

⇒ S_{15} = \(\frac{15}{2}\) [2 × 4 + (15 – 1) (- 15)]

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

⇒ S_{15} = \(\frac{15}{2}\) [8 – 70]

⇒ S_{15} = \(\frac{15}{2}\) × (- 62)

⇒ S_{15} = 15 × (- 31)

⇒ S_{15} = – 465

Hence, S_{15} = – 465.

Question 11.

If the sum of the first n terms of an AP is 4n – n^{2}, what is the first term (i.e., S4)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and 71th terms.

Solution :

We have, S_{n} = 4n – n^{2}

Putting n = 1, We get

S_{1} = 4 × 1 – (1)^{2}

⇒ S_{1} = 4 – 1

⇒ S_{1} = 3

⇒ a_{1} = 3

Putting n = 2, we get

S_{2} = 4 × 2 – (2)^{2}

⇒ S_{2} = 8 – 4

⇒ S_{2} = 4

⇒ a_{4} + a_{2} = 4

⇒ 3 + a_{2} = 4

⇒ a_{2} = 4 – 3

⇒ a_{2} = 1

Now, d = a_{2} – a_{1}

= 1 – 3 = – 2

a_{3} = 3 + (3 – 1) × (- 2)

[∵ a_{n} = a + (n – 1)d]

a_{3} = 3 + 2 × (- 2)

a_{3} = 3 – 4

a_{3} = – 1

a_{10} = 3 + (10 – 1) × (- 2)

a_{10} = 3 + 9 × (- 2)

a_{10} = 3 – 18 = -15

And a_{n} = 3 + (n – 1) × (- 2)

a_{n} = 3 – 2n + 2

a_{n} = 5 – 2n

Hence a_{1} = 3, S_{2} = 4, a_{2} = 1, a_{3} = – 1, a_{10} = – 15, a_{n} = 56 – 2n.

Question 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, …, 240.

Since, each number is divisible by 6.

So it is an AP in which

a = 6,

d = a_{2} – a_{1}

= 12 – 6 = 6,

n = 40 and l = 240

Now, S_{40} = [16 + 240]

[∵ S_{n} = (a + l)]

S_{40} = 20 × 246

S_{40} = 4920

Hence, the sum of the first 40 positive integers divisible by 6 = 4920.

Question 13.

Find the sum of the first 15 multiples of 8.

Solution :

The first 15 multiples of 8 are 8, 16, 24, 32, …, 120.

Since each number is multiple of 8, So, it is an AP in which

a = 8,d = a_{2} – a_{1} = 16 – 8 = 8, n = 5 and l = 120

Now, S_{15} – a_{1} = \(\frac{15}{2}\) [8 + 120]

[∵ S_{n} = \(\frac{n}{2}\) (a + l)]

S_{15} = \(\frac{15}{2}\) × 128

S_{15} = 15 × 64

S_{15} = 960

Hence, the sum of the first 15 multiples of 8 = 960.

Question 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.

Since, difference between each two consecutive odd numbers is 2. So it is an AP.

In which a = 1, d = a_{2} – a_{1} = 3 – 1 = 2, l = a_{n} =49

a + (n – 1)d = 49[∵ a + (n – 1) d = a_{n}]

1 + (n – 1) × 2 = 49

1 + 2n – 2 = 49

2n – 1 = 49

2n = 49 + 1

2n = 50

n = \(\frac{50}{2}\)

n = 25

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

S_{25} = \(\frac{25}{2}\) [1 + 49]

S_{25} = \(\frac{25}{2}\) × 50

S_{25} = 25 × 25

S_{25} = 625

Hence, the sum of odd numbers between 0 and 50 = 625.

Question 15.

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc …, the penalty for each succeeding day being ₹ 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 :

We have,

Penalty for the first day = ₹ 200

Penalty for the second day = ₹ 250

Penalty for the third day = ₹ 300

So, the sequence of penalty is :200, 250, 300, …………..

Here, a = 200

The penalty for each succeeding day is ₹ 50

more than for each the preceding day.

So, d = 250 – 200 = 50

and n = 30

Now, S_{30} = \(\frac{30}{2}\) [2 × 200 + (30 – 1) x50]

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

S_{30} = 15[400 + 29 × 50]

S_{30} = 15 [400 + 1450]

S_{30} =15 × 1850

S_{30} = 27750

Hence, the penalty paid for 30 days = ₹ 27750.

Question 16.

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

Solution :

Let Ist Prize be ₹ a

Then 2nd prize = ₹ (a – 20)

3rd prize = ₹ (a – 20 – 20) = (a – 40)

Thus, the seven prize are a,

(a – 20), (a – 40), …………….., (a – 120)

Since, each prize is Rs. 20 less than its pre ceding prize. So it is an AP.

Here first term (a) = a

Common difference (d)

= a_{2} – a_{1}

= a – 20 – a = – 20

Number of prizes (n)= 7 and

a_{n} = l = a – 120

Sum of prizes (S_{n}) = 700

Now, S_{7} = 700

⇒ l [a + a – 120] = 700

[∵ S_{n} = \(\frac{n}{2}\) [a + l]

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

⇒ 2a – 120 = 200

⇒ 2a = 200 + 120 = 320

⇒ a = \(\frac{320}{2}\)

⇒ a = 160

So, given prizes are 160, 160 – 20, 160 – 40, 160 – 60, 160 – 80, 160 – 100, 160 – 120,

i.e. 160, 140, 120, 100, 80, 60, 40.

Hence, seven prizes are ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, ₹ 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 :

Trees planted by three sections of class I = 3 × 1 = 3

Trees planted by three sections of Class II = 3 × 2 = 6

Trees planted by three sections of Class III = 3× 3 = 9

Trees planted by three sections of class IV = 3 × 4 = 12

Trees planted by three section of class XII = 3 × 12 = 36

. .

. .

. .

. .

. .

∴ The sequence of trees is : 3, 6, 9, 12, …, 36

a_{2} – a_{1} = 6 – 3 = 3

a_{3} – a_{2} = 9 – 6 = 3

a_{4} – a_{3} = 1293

∵ a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3}

The given sequence of trees forms an AP.

Here a = 3, d = a_{2} – a_{1} = 6 – 3 = 3,

a_{n} = l = 36, n = 12

Total number of trees will be planted = S_{n}

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

= \(\frac{12}{2}\) [3 + 36]

= 6 × 39 = 234

Hence, the total number of trees will be planted = 234.

Question 18.

A spiral is made up of succes¬sive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm … as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = \(\frac{22}{7}\))

Solution:

Let l_{1}, l_{2}, l_{3}, l_{4}, ……………….., l_{13} be the lengths (Circumferences) of semicircles of radii r_{1} = 0.5 cm, r_{2} = 1.0 cm, r_{3} = 1.5 cm, r_{4} = 2.0 cm,… respectively.

Then

l_{1} = πr_{1}

= π × 0.5

= \(\frac{\pi}{2} \mathrm{~cm}=\frac{22}{14} \mathrm{~cm}=\frac{11}{7}\) cm

l_{2} = πr_{2}

= π × 1.0

= π cm = \(\frac{22}{7}\) cm

l_{3} = πr_{3}

= π × 1.5

= 1.5π cm = 1.5 × \(\frac{22}{7}\) = \(\frac{33}{7}\) cm

l_{4} = πr_{4}

= π × 2.0

= 2π cm = \(\frac{44}{7}\) cm

. .

. .

. .

. .

. .

l_{13} = πr_{13}

= π × 6.5

= 6.5π = 6.5 × \(\frac{22}{7}\)

= \(\frac{143}{7}\)

The sequence of length of spiral is \(\frac{11}{7}, \frac{22}{7}, \frac{33}{7}, \frac{44}{7}, \ldots, \frac{143}{7}\).

a_{2} – a_{1} = \(\frac{22}{7}-\frac{11}{7}=\frac{11}{7}\)

a_{3} – a_{2} = \(\frac{33}{7}-\frac{22}{7}=\frac{11}{7}\)

a_{4} – a_{3} = \(\frac{44}{7}-\frac{33}{7}=\frac{11}{7}\)

∵ a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3} =

The sequence of length of spiral forms an

Here a = \(\frac{11}{7}\)

d = a_{2} – a_{1}

= \(\frac{22}{7}-\frac{11}{7}=\frac{11}{7}\)

l = \(\frac{143}{7}\),

and n = 13

Now, the length of such a spiral made up of thirteen consecutive semicircles = S_{13}

S_{13} = \(\frac{13}{2}\left[\frac{11}{7}+\frac{143}{7}\right]\)

[∵ S_{n} = \(\frac{n}{2}\) [a + l]]

⇒ S_{13} = \(\frac{13}{2}\left[\frac{154}{7}\right]\)

⇒ S_{13} = \(\frac{13}{2}\) × 22

⇒ S_{13} = 13 × 11

⇒ S_{13} = 143.

Hence, the length of a spiral made up of thir¬teen consecutive semicircles = 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 to it and so on (see in figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution :

Sequence of logs row-wise is as follows :

20, 19, 18, 17, ………………

a_{2} – a_{1} = 19 – 20 = – 1

a_{3} – a_{2} = 18 – 19 = – 1

a_{4} – a_{3} = 17 – 18 = – 1

∵ a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3}

∴ Sequence of logs row-wise forms an AP.

Here, a = 20, d = a_{2} – a_{1} = 19 – 20 = – 1, S_{n} = 200

\(\frac{n}{2}\) [2a + (n – 1)d] = 200

n[2 × 20 + (n – 1) × (- 1)] = 200 × 2

n[40 – n + 1] = 400

n [41 – n] = 400

41n – n^{2} = 400

n^{2} – 41n + 400 = 0

n^{2} – (25 + 16)n + 400 = 0

[1 × 400 = 400

25 × 16 = 400

25 + 16 = 41]

n^{2} – 25n – 16n + 400 = 0

⇒ n(n – 25) – 16(n – 25) = 0

⇒(n – 25) (n – 16) = 0

⇒ n – 25 = 0 or n – 16 = 0

⇒ n = 25 or n = 16

If n = 25

Number of logs in top row = a_{25}

= 20 + (25 – 1) × (- 1) [∵ a_{n} = a + (n – 1)d]

= 20 – 24 = – 4

– 4 is not meaningful

So, we reject n = 25.

∴ n = 16

∴ Number of logs in top row = a_{16}

= 20 + (16 – 1) × (- 1)

= 20 – 15 = 5

Hence, the number of rows = 16 and

the number of logs in top row = 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 line. There are ten potatoes in the line (see figure).

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 draped in the bucket. What is the total distance the competitor has to run ?

Solution :

Distance run to pick up the 1st potato = 2 × 5 = 10 m

Distance run to pick up the 2nd potato = 2 × (5 + 3) = 16 m

Distance run to pick up the 3rd potato = 2 × (5 + 3 + 3) = 22 m

Distance run to pick up the 4th potato = 2 × (5 + 3 + 3 + 3) = 28 m

The sequence of distances is 10, 16, 22, 28.

a_{2} – a_{1} = 16 – 10 = 6

a_{3} – a_{2} – 22 – 16 = 6

a_{4} – a_{3} = 28 – 22 = 6

∵ a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3}

The sequence forms an AP.

Here, a = 10, d = a_{2} – a_{1} = 16 – 10 = 6, n = 10

Now, total distance run by a competitor = S_{10}

= \(\frac{10}{2}\) [2 × 10 + (10 – 1) × 61]

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

= 5 [20 + 9 × 6]

= 5 [20 + 54]

= 5[74] = 370 m

Hence, the total distance run by competitor = 370 m.