RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 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) The given AP is :
2, 7, 12, …..
Here a = 2, d = 7 – 2 = 5
and n = 10
Using the formula
S_n = \(\frac{n}{2}\) [2a + (n- 1 )d]
∴ S₁₀ = \(\frac{10}{2}\)[2 × 2 + (10 -1) 5]
= 5 [4 + 45]
= 5 × 49 = 245
Hence, sum of 10 terms = 245

(ii) The given AP is: – 37, – 33, – 29, ….
Here a = – 37, d = – 33 + 37 = 4
and n = 12
Using the formula
S_n = \(\frac{n}{2}\)[2a + (n- 1 )d]
∴ S₁₂ = \(\frac{12}{2}\)[2 × (-37) + (12 – 1)4]
= 6 [-74 + 44]
= 6 × (-30) = – 180
Hence, sum of 12 terms = – 180

(iii) The given AP is : 0.6, 1.7, 2.8, ….
Here a = 0.6, d= 1.7 – 0.6 = 1.1
and n = 100
Using the formula
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₀₀ = \(\frac{100}{2}\)[2 (0.6) + (100 – 1) 1.1]
= 50 [1.2 + 108.9]
= 50 × 110.1 = 5505
Hence sum of 100 terms = 5505

(iv) The given AP is :\(\frac{1}{15}\), \(\frac{1}{12}\), \(\frac{1}{10}\), ……….


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

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) The given AP is :

(ii) The given AP is : 34 + 32 + 30 + + 10
Here, a = 34, d = 32 – 34 = -2
and l = a_n = 10
⇒ a + (n – 1) d = 10
or 34 + (n – 1) (-2) = 10
or -2 (n – 1) = 10 – 34 = -24
or n – 1 = \(\frac{24}{2}\) = 12
or n = 12 + 1 = 13
Now, S₁₃ = \(\frac{13}{2}\)[ 34 + 10]
[∵ S_n = \(\frac{n}{2}\)[a + l]]
= \(\frac{13}{2}\) × 44 = 286
So, 34 + 32 + 30 + …. + 10 = 286

(iii) The given AP is :
– 5 + (- 8) + (- 11) + ….. +(-230)
Here, a = -5, d = -8 + 5 = -3
and l = a_n = -230
⇒ a + (n – 1) d = -230
or – 5 + (n – 1) (- 3) = -230
or – 3 (n – 1) = -230 + 5 = -225
or n- 1 = \(\frac{225}{3}\) = 75
or n = 75 + 1 = 76
Now, S₇₆ = \(\frac{76}{2}\)[-5 + (-230)]
[∵ S_n = \(\frac{n}{2}\)[a + l]]
Hence – 5 + (- 8) + (- 11) + …. + (- 230) = -8930

Question 3.
In an AP :
(i) given a = 5, d = 3, a_n = 50, find n and S_n.
(ii) given a = 7, a₁₃ = 35, find d and S₁₃.
(iii) given a₁₂ = 37, d= 3, find a and S₁₂.
(iv) given a₃ = 15, S₁₀ = 125, find d and a₁₀.
(v) given d = 5, S₉ = 75, find a and a₉.
(vi) given a = 2, a_n = 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 there are total 9 terms. Find a.
Solution:
(i) Given a = 5, d = 3, a_n = 50
∵ a_n = 50
∴ a + (n – 1) d = 50
or 5 + (n – 1) 3 = 50
or 3 (n – 1) = 50 – 5 = 45
or n – 1 = = 15
⇒ n = 15 + 1 = 16
Now, S_n = \(\frac{n}{2}\)[a + l]
= \(\frac{16}{2}\)[5 + 50] = 8 55
= 440
Hence, n = 16 and S_n= 440

(ii) Given a = 7, a₁₃ = 35
∵ a₁₃ = 35
⇒ a + (n – 1) d = 35
or 7+ (13 – 1)d = 35
or 12d = 35 – 7 = 28
or d = \(\frac{28}{12}\) = \(\frac{7}{3}\)
Now, S₁₃ = \(\frac{13}{2}\)[7 + 35]
[∵ S_n = \(\frac{n}{2}\)[a + l]]
= \(\frac{13}{2}\) × 42 = 13 × 21 = 273
Hence, d = \(\frac{7}{3}\) and S₁₃ = 273

(iii) Given a₁₂ = 37, d = 3
∵ a₁₂ = 37
⇒ a + (n – 1) d = 37
or a + (12 – 1)3 =37
a + 11 × 3 = 37
or a = 37 – 33 = 4
Now, S₁₂ = \(\frac{12}{2}\)[4 + 37]
[∵ S_n = \(\frac{n}{2}\)[a + l]]
= 6 × 41 = 246
Hence, a = 4 and S₁₂ = 246

(iv) Given a₃ = 15, S₁₀ = 125
∵ a₃ = 15
∴ a + (n – 1) d = 35
or a + (3 – 1) d = 15
or a + 2d = 15 ….(1)
\(\frac{10}{2}\)[2a + (10 – 1) d] = 125
[∵ S_n = \(\frac{n}{2}\)[2a + (n – 1)d]]
or 5 [2a + 9d] = 125
or 2a + 9d = \(\frac{125}{5}\) = 25
or 2a + 9d = 25 ….(2)
From (1) a = 15 – 2d ….(3)
Substituting the value of a in (2)
2(15 – 2d) + 9d = 25
or 30 – 4d + 9d = 25
or 5d = 25 – 30
or d = \(\frac{-5}{5}\) = -1
Substituting the value of d in (3)
a = 15 – 2 (-1)
a = 15 + 2= 17
Now, a₁₀ = 17 + (10 – 1)(-1)
[∵ a_n = a + (n – 1)d]
= 17 – 9 = 8
So, d = – 1 and a₁₀ = 8

(v) Given d = 5, S₉ = 75
∵ S₉ = 75
\(\frac{9}{2}\)[2a + (9 – 1)5] = 75
[∵ S_n = \(\frac{n}{2}\)[2a + (n – 1 )d]

(vi) Given a = 2, d = 8, S_n = 90
∵ S_n = 90
∴ \(\frac{n}{2}\)[2a + (n – 1) d] = 90
or \(\frac{n}{2}\)[2 × 2 +(n – 1)8] = 90
or n [2 + 4n – 4] = 90
or n (4n – 2) = 90
or 4n² – 2n – 90 = 0
or 2n² – n – 45 = 0
or 2n² – 10n + 9n – 45 = 0 | S = -2
P = -45 × 2 = -90
or 2n[n – 5] + 9(n – 5) = 0
or (2n + 9) (n – 5) = 0
i.e., 2n + 9 = 0
or n – 5 = 0
i.e., n = \(-\frac{9}{2}\) or n = 5
∵ n cannot be negative therefore ignoring
n = \(-\frac{9}{2}\)
∴ n = 5
Now, a_n = a₅ = a + (n -1)d
= 2 + (5 – 1) 8
= 2 + 32 = 34
So, n = 5 and a_n= 34

(vii) Given a = 8, a_n = 62, S_n = 210
∵ S_n = 210
∴ \(\frac{n}{2}\)[a + a_n] = 210
or \(\frac{n}{2}\)[8 + 62] = 210
or \(\frac{n}{2}\) × 70 = 210
or n = \(\frac{210}{35}\) = 6
Now, a_n = 62
8 + (6 – 1) of = 62 [∵ a_n = a + {n – 1)d]
or 5d = 62 – 8 = 54
or d = \(\frac{54}{5}\)
or n = 6 and d = \(\frac{54}{5}\)

(viii) Given a_n = 4, d = 2, S_n = – 14
∵ a_n = 4
∴ a + (n – 1) d = 4
or a + (n – 1)2 =4
or a + 2n – 2 =4
or a = 6 – 2n ….(1)
and S_n = -14
or \(\frac{n}{2}\)[a + a_n] = – 14
or \(\frac{n}{2}\)[6 – 2n + 4] = – 14 [Using (1)]
or \(\frac{n}{2}\)[10 – 2n] = -14
or 5n – n² + 14 = 10
or n² – 5n – 14 = 0 |S = – 5
or n² – 7n + 2n – 14 = 0 | P = 1 × -14 = -14
or n² – 7n + 2n – 14 = 0
or n(n – 7) + 2 (n – 7) = 0
or (n – 7) (n + 2) = 0
i.e., n – 7 = 0 or n + 2 = 0
n = 7 or n = -2
∵ n cannot be negative
∴ ignoring n = – 2
∴ n = 7
Substituting the value of n in (1)
a = 6 – 2 × 7
a = 6 – 14 = – 8
So, a = – 8 and n = 7

(ix) Given a = 3, n = 8, S = 192
∵ S = 192
⇒ S₈ = 192 [∵ n = 8]
or \(\frac{8}{2}\)[2 × 3 + (8 – 1)d] = 192
[∵ S_n = \(\frac{n}{2}\)[2a + (n – 1)d]]
or 4 [6 + 7d] = 192
or 6 + 7 d = \(\frac{192}{4}\) = 48
or 7d = 48 – 6 = 42
or d = \(\frac{42}{7}\) = 6

(x) Given l = 28, S = 144
and there are 9 terms in all.
∴ n = 9, l = a₉ = 28, S₉ = 144
∵ a₉ = 28
or a + (9 – 1) d = 28
[∵ a_n = a_n = 9 + (n – 1)d]
or a + 8d = 28 …(1)
and S₉ = 144
⇒ \(\frac{9}{2}\)[a + 28]= 144 [∵ S_n = \(\frac{n}{2}\)[a + a_n]]
or a + 28 = \(\frac{144 \times 2}{9}\) = 32
⇒ a = 32 – 28 = 4

Question 4.
How many terms of the AP : 9, 17, 25,….. must be taken to give a sum of 636?
Solution:
Given : AP 9, 17, 25, …..
Here, a = 9, d = 17 – 9 = 8
Since, S_n = 636
∴ \(\frac{n}{2}\)[2o + (n – 1) d] = 636
or \(\frac{n}{2}\)[2 (9) + (n – 1)8] = 636
or \(\frac{n}{2}\)[18 + 8n – 8] = 636
or n [4n + 5] = 636
or 4n² + 5n – 636 = 0
or a = 4, b = 5, c = – 636
D = (5)² – 4 × 4 × (-636)
= 25 + 10176 = 10201

∵ n cannot be negative
So, ignoring n = \(-\frac{53}{4}\)
∴ n = 12
Hence, the sum of 12 terms of the given AP is 636.

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: Given a = 5; l = a_n = 45
and S_n = 400
∵ a_n = 45
∴ a + (n – 1) d = 45
or 5 + (n – 1)d = 45
or (n – 1)d = 45 -5 = 40
and S_n = 400
\(\frac{n}{2}\)[a + a_n] = 400
or \(\frac{n}{2}\)[5 + 45] = 400
or 25 n = 400
or n = \(\frac{400}{25}\) = 16
Substituting the value of n in (1)
(16 – 1)d = 40
or 15d = 40
or d = \(\frac{40}{15}\) = \(\frac{8}{3}\)
So, n = 16 and d = \(\frac{8}{32}\)

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:
Given that
First term = a= 17;
Last term = l = a_n = 350
and common difference = d = 9
∵ l = a_n = 350
∴ a + (n – 1)d = 350
17 + (n – 1)9 = 350
or 9(n – 1) = 350 – 17 = 333
or n – 1 = \(\frac{333}{9}\) = 37
or n = 37 + 1 = 38
Now, S₃₈ = \(\frac{n}{2}\)[a + l]
= \(\frac{38}{2}\)[17 + 350]
= 19 × 367 = 6973
Hence, the sum of 38 terms of the given AP is 6973.

Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
Given that d = 7, a₂₂ =149
and n =22
∵ a₂₂ = 149
∴ a + (n – 1) d = 149
or a + (22 – 1)7 = 149
or a + 147 = 149
or a = 149 – 147 = 2
Now, S₂₂ = \(\frac{n}{2}\)[a + a₂₂]
= \(\frac{22}{11}\)[2 + 149]
= 11 × 151 = 1661
Hence, the sum of first 22 terms of the given AP is 1661.

Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Let ‘a’ and d’ be the first term and the common difference respectively.
Given that a₂ = 14; a₃ = 18
and n = 51
∵ a₂ = 14
∴ a + (2 – 1)d = 14
or a + d = 14
⇒ a = 14 – d ….(1)
and a₃ = 18 (Given)
∴ a + (3 – 1)d = 18
or a + 2d = 18
or 14 – d + 2d = 18
or d = 18 – 14 = 4
or d = 4
Substituting the value of d in (1),
a = 14 – 4 = 10
Now, S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S_n = \(\frac{51}{2}\)[2 × 10 + (51 – 1)4]
= \(\frac{51}{2}\)[20 + 200]
= \(\frac{51}{2}\) × 220 = 51 × 110
= 5610
Hence, the sum of first 51 terms of the given AP is 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’ and ‘d’ be the first term and the common difference of the given AP respectively.
According to the first condition,
S₇ = 49
∴ \(\frac{n}{2}\)[2a + (n – 1)d] = S_n
or \(\frac{7}{2}\)[2a + (7 – 1)d] = S₇
or \(\frac{7}{2}\)[2a + 6d] = 49
or a + 3d = 7
or a = 7 – 3d …(1)
According to the second condition,
S₁₇ = 289
∴ \(\frac{17}{2}\)[2a + (17 – 1)d] = 289
[∵ S_n = \(\frac{n}{2}\)[2a + (n – 1)d]]
\(\frac{1}{2}\)[2a + 16d] = 289
⇒ a + 8d = \(\frac{289}{17}\) = 17
Substituting the value of a from (1)
7 – 3d+ 8d = 17
5d = 17 – 7 = 10
d = \(\frac{10}{5}\) = 2
Substituting the value of d in (1)
a = 7 – 3 × 2
⇒ a = 7 – 6 = 1
Now, S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
= \(\frac{n}{2}\)[2 × 1 + (n – 1) × 2]
= \(\frac{n}{2}\)[2 + 2n – 2] \(\frac{n}{2}\) × 2n = n²
Hence, the sum of the first n terms of the given AP is n².

Question 10.
Show that a₁, a₂, ….., 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) Given that a_n = 3 + 4n ….(1)
Substituting the different values of n in (1)
a₁ = 3 + 4(1) = 7;
a₂ = 3 + 4(2) = 11
a₃ = 3 + 4(3) = 15,
Now we shall find the values of a₂ – a₁, a₃ – a₂
∵ a₂ – a₁ = 11 – 7 = 4
and a₃ – a₂ = 4 = d (say)
∴ The given sequence is in the form of AP
Here, a = 7, d = 4 and n = 15
∴ S₁₅ = \(\frac{n}{2}\)[2a + (n – 1)d]
= \(\frac{15}{2}\)[2(7) + (15 – 1)4]
= \(\frac{15}{2}\)[14 + 56] = \(\frac{15}{2}\) × 70
= 15 × 35 = 525

(ii) Given that a_n = 9 – 5n …(1)
Substituting the different values of n in (1)
a₁ = 9 – 5(1) = 4;
a₂ = 9 – 5 (2) = -1;
a₃ = 9 – 5 (3) = -6
Now, a₂ – a₁ = -1 – 4 = – 5
and a₃ – a₂ = -6 + 1 = -5
∵ a₂ – a₁ = a₃ – a₂ = – 5 = d (say)
∴ The given sequence is in the form of AP
Here a = 4, d = – 5 and n = 15
∴ S₁₅ = \(\frac{n}{2}\)[2a + (n – 1)d]
= \(\frac{15}{2}\)[2(4) + (15 – 1)(-5)]
= \(\frac{15}{2}\)[8 – 70]
= \(\frac{15}{2}\)(-62)
= 15 × (-31) = -465
Hence, sequence = – 4, – 1,-6, …. and sum = – 465

Question 11.
If the sum of the first n terms of an AP is 4n – n², what is the first term (that is Sj)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
Given that the sum of the n terms of the AP is
S_n = 4n – n² ……..(1)
Substituting the value n = 1 in (1)
S₁ = 4(1) – (1)² = 4 – 1
⇒ S₁ = 3
∴ a = a₁ = S₁ = 3
Substituting the value n = 2 in (1)
S₂ = 4(2) – (2)² = 8 – 4
⇒ S₂ = 4
or a₁ + a₂ = 4
or 3 + a₂ = 4
or a₂ = 4 – 3 = 1
Substituting the value n = 3 in (1)
S₃ = 4 (3) – (3)2 = 12 – 9
⇒ S₃ = 3
or S₂ + a₃ = 3
or 4 + a₃ = 3
or a₃ = 3 – 4 = -1
or d = a₂ – a₁
= 1 – 3 = – 2
∴ a₁₀ = a + (n – 1 )d
= 3 + (10 – 1) (-2)
⇒ a₁₀ = 3 – 18 = – 15
and a_n = a + (n – 1) d
= 3 + (n – 1) (- 2)
= 3 – 2n + 2
⇒ a_n = 5 – 2n
So S₁ = 3
Sum of the first two terms S₂ = 4
Second term a₂ = 1
Third term a₃ = -1
Tenth term a₁₀ = -15
and n^th term a_n = 5 – 2n

Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
Positive integers divisible by 6 are : 6, 12, 18, 24, 30, 36, 42, …..
Here, a = a₁ = 6, a₂ = 12, a₃ = 18, a₄ = 24
a₂ – a₁ = 12 – 6 = 6
a₃ – a₂ = 18 – 12 = 6
a₄ – a₁₀ = 24 – 18 = 6
a₂ – a₁ = a₃ – a₂0
= a₄ – a₃ = 6 = d (say)
Using the formula S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
S₄₀ = \(\frac{40}{2}\)[2(6)+ (40 – 1)6]
= 20 [12 + 234]
= 20 (246) = 4920
Hence, the sum of 40 positive integers divisible by 6 is 4920.

Question 13.
Find the sum of the first 15 multiples of 8.
Solution:
The multiples of 8 are : 8, 16, 24, 32, 40, 48,….,
Here a = a₁ = 8, a₂ = 16, a₃ = 24, a₄ = 32
a₂ – a₁ = 16 – 8 = 8
a₃ – a₂ = 24 – 16 = 8
v a₂ – a₁ = a₃ – a₂ = 8 = d (say)
Using the formula S_n = \(\frac{n}{2}\)[2a + (n- 1 )d]
S₁₅ = \(\frac{15}{2}\)[2(8)+ (15 – 1)8]
= \(\frac{15}{2}\)[16 + 112]
= \(\frac{15}{2}\) × 128 = 960
Hence, the sum of the first 15 multiples of 8 is 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, 9, …. , 49
Here, a = a₁ = 1, a₂ = 3, a₃ = 5, a₄ = 7
and l = a_n = 49
a₂ – a₁ = 3 – 1 = 2
a₃ – a₂2 = 5 – 3 = 2
∵ a₂ – a₁ = a₃ – a₂ = 2 = d (say)
Also l = a_n = 49
a + (n – 1) d = 49
1 +(n – 1)2 = 49
or 2 (n – 1) = 49 – 1 = 48
or n – 1 = \(\frac{48}{2}\) = 24
or n = 24 + 1 = 25
Using the formula S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
S₂₅ = \(\frac{25}{2}\)[2(1) + (25 – 1)2]
= \(\frac{25}{2}\)[2 + 48]
= \(\frac{25}{2}\) × 50 = 625
Hence, the sum of the odd numbers between 0 and 50 is 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:
The penalty for the first day, the second day and the third day are
₹ 200, ₹ 250, ₹ 300 respectively
Now, the penalty for each succeeding day goes on increasing by ₹ 50.
∴ The required AP is : ₹ 200, ₹ 250, ₹ 300, ₹ 350,
Here, a = a₁ = 200; d = 50 and n = 30
The amount of penalty to be given after 30 days
= S₃₀ = \(\frac{n}{2}\)[2a + (n- 1 )d]
= Y [2(200) + (30 – 1)50]
= 15 [400+ 1450]
= 15 (1850) = 27750
Hence, if the contractor delays the work by 30 days, then he shall have to pay ₹ 27,750 as penalty.

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 the value of prize given to the first student = ₹ x
Then, the value of prize given to the second student = ₹ (x – 20)
and, the value of prize given to the third student = ₹ (x – 20 – 20)
= ₹ (x – 40)
The required sequence is : ₹ x, ₹ (x – 20), ₹(x – 40),…..
Which form an AP, in which
a = ₹ x, d = -20 and n = 7
Using the formula
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
S₇ = \(\frac{7}{2}\)[2(x) + (7 – 1) (- 20)]
S₇ = \(\frac{7}{2}\)[2x – 120] = 7 (x – 60)
According to the question,
7 (x – 60) = 700
⇒ x – 60 = \(\frac{700}{7}\) = 100
⇒ x = 100 + 60
⇒ x = 160
Hence, the 7 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:
Number of trees planted by three sections of class I = 3 × 1 = 3
Number of trees planted by three sections of class II = 3 × 2 = 6
Number of trees planted by three sections of class III = 3 × 3 = 9
…………………………………………………..
…………………………………………………..
…………………………………………………..
Number of trees planted by three sections of class XII = 3 × 12 = 36
The required AP is : 3, 6, 9, ……… , 36
Here, a = a₁ = 3; a₂ = 6; a₃ = 9
and l = a_n = 36; n = 12
d = a₂ – a₁ = 6 – 3 = 3
Total number of trees planted by students = S₁₂
Number of trees planted by students n r „
= \(\frac{n}{2}\)[a + l]
= \(\frac{12}{1}\)[3 + 36]
= 6 × 39 = 234
Hence, to reduce air pollution, 234 trees will be planted by the students.

Question 18.
A spiral is made up of successive 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 Fig. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = \(\frac{22}{7}\))

Hint : Length of successive semicircles is l₁, l₂, l₃, l₄, with centres at A, B, A, B, …..respectively.]
Solution:
Let
l₁ = length of first semicircle
= πr₁ = π(0.5) = \(\frac{\pi}{2}\)
l₂ = length of second semicircle
= πr₂ = π(1) = π
l₃ = length of third semicircle
= πr₃ = π( 1.5) = \(\frac{3 \pi}{2}\)
and l₄ = length of fourth semicircle
= πr₄ = π(2)
= 2π and so on so forth.
∵ Length of each successive semicircles form an AP

Hence, the total length of spiral made up of thirteen consecutive semicircles is 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 Fig.). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:
Number of logs in the bottom row = 20
Number of logs in the second row = 19
Number of logs in the third row = 18
and so on so forth.
∴ Number of logs in each row form an AP
Here, a = a₁ = 20;
a₂ = 19; a₃ = 18
d = a₂ – a₁
= 19 – 20 = – 1
Let S_n represent the total number of logs. Using the formula,
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S_n = \(\frac{n}{2}\)[2(20) + (n- 1)(-1)]
= \(\frac{n}{2}\)[40 – n + 1] = \(\frac{n}{2}\)[41 – n]
According to the question
\(\frac{n}{1}\)[41 – n] = 200
or 41n – n² = 400
or n² – 41n + 400 = 0 | S = 41
or n² – 16n – 25n + 400 = 0 | P = 400
or n(n – 16) – 25(n – 16) = 0
or (n -16) (n – 25) = 0
i.e., n – 16 = 0 or n – 25 = 0
i.e., n = 16 or n = 25
n = 16, 25
Case I: when n = 25
a₂₅ = a + (n – 1)d
= 20 + (25 – 1) (-1)
= 20 – 24 = -4
which is impossible
∴ We ignore n = 25

Case II: When n = 16
a₁₆ = a + (n – 1)d
= 20 + (16 – 1)(-1)
= 20 – 15 = 5
Hence, there are 16 rows in all and there are 5 logs in the top row.

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 Fig).

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 iji 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 × 5 + 2 × (5 + 3)]
Solution:
Distance run to pick up the first potato = 2 (5) m = 10 m
Distance between consecutive potatoes = 3 m
∴ Distance run to pick up the second potato
= 2 (5 + 3) m = 16 m
Distance run to pick up the third potato
= 2 (5 + 3 + 3) m = 22 m
and this process continues.
It is obvious from here that in this case an AP is formed.
10m, 16 m, 22 m, 28 m, ….
Here, a = a₁ = 10; a₂ = 16; a₃ = 22,…..
d = a₂ – a₁ = 16 – 10 = 6
and n = 10
∴ The total distance the competitor has to run
= S₁₀ = \(\frac{n}{2}\)[2a + (n- 1)d]
= \(\frac{10}{2}\)[2(10) + (10 – 1)6]
= 5 [20 + 54]
5 × 74 = 370
Hence, the competitor will have to run the total distance of 370 m.