OP Malhotra Class 10 Maths Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(b)

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S Chand Class 10 ICSE Maths Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(b)

Question 1.
Find the sum of the:
(i) first 15 terms of the A.P: 2, 5, 8, 11, …
(ii) first 50 terms of the A.P: -27, -23, -19, …
(iii) first 17 terms of the A.P: = \(\frac{1}{5}, \frac{-3}{10}, \frac{-4}{5}\) …
(iv) first 24 terms of A.P: 0.6, 1.7, 2.8, …
Solution:
(i) First 15 terms of the A.P: 2, 5, 8, 11, …
Here, a = 2, d= 5 – 2 = 3, n = 15
S_n = \(\frac{n}{2}[2 a+(n-1) d]\)
S₁₅ = \(\frac{15}{2}[2 \times 2+(15-1) \times 3]\)
= \(\frac{15}{2}[4+42]\)
= \(\frac{15}{2} \times 46\) = 345

(ii) First 50 terms of the A.P: – 27, – 23, – 19, …
Here, a = – 27, d = – 23 – (- 27)
= – 23 + 27 = 4
n = 50
∴ S_n = \(\frac{n}{2}[2 a+(n-1) d]\)
= \(\frac{50}{2}[2 \times(-27)+(50-1) \times 4]\)
S₅₀ = 25[-54 + 49 x 4] = 25[- 54 + 196] = 25 x 142 = 3550

(iii) First 17 terms of the A.P: = \(\frac{1}{5}, \frac{-3}{10}, \frac{-4}{5}\) …
Here, a = \(\frac {1}{5}\), d = \(\frac { -3 }{ 10 }\) – \(\frac { 1 }{ 5 }\)
\(\frac{-3-2}{10}=\frac{-5}{10}=\frac{-1}{2}\)
and n = 1
S_n = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
S₁₇ = \(\frac{17}{2}\left[2 \times \frac{1}{5}+(17-1)\left(\frac{-1}{2}\right)\right]\)
= \(\frac{17}{2}\left[\frac{2}{5}+16\left(\frac{-1}{2}\right)\right]\)
= \(\frac{17}{2}\left[\frac{2}{5}-8\right]\)
= \(\frac{17}{2}\left(\frac{2-40}{5}\right)=\frac{17}{2} \times \frac{-38}{5}\)
= \(\frac { -323 }{ 5 }\) = – 64\(\frac { 3 }{ 5 }\)

(iv) First 24 terms of A.P: 0.6, 1.7, 2.8, …
Here, a = 0.6, d = 1.7 – 0.6 = 1.1, n = 24
∴ S_n = \(\frac{n}{2}[2 a+(n-1) d]\)
S₂₄ = \(\frac { 24 }{ 2 }\)[2 x 0.6 + (24 – 1) (1.1)]
= 12[1.2 + 23 x 1.1]
= 12[1.2 + 25.3]
= 12 x 26.5
= 318.0

Question 2.
Find the sum:
(i) 34 + 32 + 30 + … + 2
(ii) 7 + 9 \(\frac { 1 }{ 2 }\) + 12 + … + 67.
Solution:
(i) 34 + 32 + 30 + … + 2
Here, a = 34, d = 32 – 34 = – 2, l = 2
a_n = a + (n – 1)d
⇒ 2 = 34 + (n – 1) x (- 2)
⇒ 2 – 34 = – 2(n – 1)
⇒ \(\frac { -32 }{ -2 }\) = n – 1
⇒ n – 1 = 16
n = 16 + 1 = 17
∴ S_n = \(\frac{n}{2}(a+l)=\frac{17}{2}(34+2)\)
= \(\frac { 17 }{ 2 }\) x 36 = 306

(ii) 7 + 9 \(\frac { 1 }{ 2 }\) + 12 + … + 67
Here, a = 7, d = 9\(\frac { 1 }{ 2 }\) – 7 = 2\(\frac { 1 }{ 2 }\) = \(\frac { 5 }{ 2 }\), l = 67
l = (a_n) = a + (n – 1 )d
⇒ 67 = 7 + (n – 1)\(\frac { 1 }{ 2 }\)
67 – 7 = \(\frac { 5 }{ 2 }\)(n – 1) ⇒\(\frac { 60×2 }{ 5 }\) n = 1
⇒ n – 1 = 24 ⇒ n = 24 + 1 = 25
∴ S₂₅ = \(\frac { n }{ 2 }\)[a + l] = \(\frac { 25 }{ 2 }\)[7 + 67]
= \(\frac { 25 }{ 2 }\) x 74 = 925

Question 3.
The common difference of an A.P. is – 2. Find its sum, if its first term is 100 and the last term is – 10.
Solution:
In an AP
d = – 2, a = 100, l = – 10
l = (a_n) = a + (n – 1 )d
– 10 = 100 + (n – 1)(- 2)
(n – 1)(- 2) = – 10 – 100 = – 110
n – 1 = \(\frac { -110 }{ – 2 }\) = 55
⇒ n = 55 + 1 = 56
∴ S₅₆ = \(\frac { n }{ 2 }\)[a + l] = \(\frac { 56 }{ 2 }\) [100 – 10]
= 28 x 90 = 2520

Question 4.
How many terms of the A.P. 54, 51, 48, … should be taken so that their sum is 513?
Solution:
AP is 54, 51, 48, … and S_n = 513
Here, a = 54, d = 51 – 54 = – 3
S_n = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
513 = \(\frac { n }{ 2 }\)[2 x 54 + (n – 1) x (- 3)]
⇒ 513 x 2 = n[108 – 3n + 3]
⇒ 1026 = 108n – 3n² + 3n
⇒ 3n² – 111n + 1026 = 0
⇒ n² – 37n + 342 = 0
⇒ n² – 18n – 19n + 342 = 0

⇒ n(n – 18) – 19(n – 18) = 0
⇒ (n – 18)(n – 19) = 0
Either n – 18 = 0, then n = 18
or n – 19 = 0, then n = 19
∴ Number of terms = 18 or 19

Question 5.
If the sum of first 9 terms of an A.P. is 72 and the common difference is 5, find the first term and the 10th term of the A.P.
Solution:
S₉ = 72, d = 5
Let a be the first term,
n = 9
∴ S₉ = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
72 = \(\frac { 9 }{ 2 }\)[2a + (9 – 1) x 5]
\(\frac { 72×2 }{ 2 }\) = 2a + 40
⇒ 16 = 2a+ 40
2a =16-40 = – 24
a = \(\frac { -24 }{ 2 }\) = -12
∴ a = – 12
a₁₀ = a + (n – 1 )d
= – 12 + (10 – 1) x 5
= – 12 + 45
= – 33

Question 6.
The sum of the first six terms of an A.P. is 42. The ratio of its 10th term to its 30th term is 1 : 3. Calculate the first and thirteenth term of the AP.
Solution:
In an AP,
Sum of first 6 terms = 42
a₁₀ : a₃₀ = 1 : 3
Let a be the first term and d be the common difference, then
S₆ = \(\frac { 6 }{ 2 }\) [2a + (K – 1)d]
42 = 3(2a + 5 d) = 6 a + 15 d
⇒ 6a + 15d = 42 … (i)
a₁₀ : a₃₀ = 1 : 3
\(\frac{a+(10-1) d}{a+(30-1) d}=\frac{1}{3} \Rightarrow \frac{a+9 d}{a+29 d}=\frac{1}{3}\)
3a + 27d = a + 29 d
3a – a = 29d – 27d
⇒ 2a = 2d
⇒ a = d
From (i)
6a + 15a = 42 ⇒ 21a = 42
⇒ a = \(\frac { 42 }{ 2 }\) = 2
∴ a = 2, d = 2
∴ First term = 2
a₁₃ = a + (n – 1)d = 2 + (13 – 1) x 2
= 2 + 12 x 2 = 2 + 24 = 26

Question 7.
The 13th term of an A.P. is four times its 3rd term. If its fifth term is 16, then find the sum of its first ten terms.
Solution:
In an AP,
a₁₃ = 4 x a₃
a₅ = 16
Let a be the first term and d be the common difference.
∴ a₅ = a + (n – 1)d = a + (5 – 1)d = a + 4d
∴ a + 4d = 16 … (i)
Similarly,
a₁₃ = a + 12d and a₃ = a + 2d
∴ a + 12d = 4 x (a + 2d)
a + 12d = 4a + 8d
12d – 8d = 4a – a ⇒ 3a = 4d
a = \(\frac { 4 }{ 3 }\)d
From (i)
\(\frac { 4 }{ 3 }\) + 4d = 16 ⇒ \(\frac { 16 }{ 3 }\)d = 16
⇒ d = \(\frac { 16×3 }{ 16 }\) = 3
∴ d = 3
and a = \(\frac { 4 }{ 3 }\) d = \(\frac { 4 }{ 3 }\) x 3 = 4
a – 4, d = 3
Now, S₁₀ = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
= \(\frac { 10 }{ 2 }\)[2 x 4+ (10 – 1) x 3] = 5(8 + 27)
= 5 x 35 = 175

Question 8.
Find the 60th term of the AP 8,16,12,… if it has a total of 60 terms. Hence find the sum of its last 10 terms.
Solution:
AP is 8, 10, 12, …
Here a = 8, d = 10 – 8 = 2, n = 60
∴ T_n = a +(n – 1)d
⇒ 60 = 8 + (n – 1) x 2
⇒ (n – 1) x 2 = 60 – 8 = 52
n – 1 = \(\frac { 52 }{ 2 }\) = 26
n = 26 + 1 = 27
T₆₀ = a + (60 – 1)d = 8 + 59 x 2
= 8 + 61 = 69
Sum of last 10 terms = S₆₀ – S₅₀
= \(\frac { 60 }{ 2 }\) (2a + 59d) – \(\frac { 50 }{ 2 }\)(2a + 49d)
= 30(2a + 59d) – 25(2a + 49d)
= 60a + 1770d – 50a – 1225d
= 10a + 545d = 10 x 8 + 545 x 2
= 80 + 1090 = 1170

Question 9.
In an A.P. if the 12th term is -13 and the sum of its first four terms is 24, find the sum of its first 10 terms.
Solution:
In an AP,
a₁₂ or T₁₂ = – 13
S₂₄ = 24
Let a be the first term and d be the common difference, then
a₁₂ = a + (n – 1 )d
⇒ – 13 = a + (12 – 1)d
⇒ a + 11d = -13 ⇒ a = – 13 – 11d
S₄ = \(\frac { n }{ 2 }\)(2a + (n – 1)d]
24 = \(\frac { 4 }{ 2 }\)[2a + 3d] = 2[2 x (- 13 – 11d) + 3d]
\(\frac { 24 }{ 2 }\)= – 26 – 22d + 3d ⇒ 12 = – 26 – 19d
⇒ 12 + 26 = – 19d ⇒ – 19d = 38
d = \(\frac { 38 }{ -19 }\) = – 2
and a = – 13 – 11d = – 13 + 11 x 2
= – 13 + 22 = 9
∴ a = 9, d = – 2
Now, S₁₀ = \(\frac { n }{ 2 }\)(2a + (n – 1)d]
= \(\frac { 10 }{ 2 }\)[2 x 9 + (10 – 1)(- 2)]
= 5[18 + 9(-2)] = 5(18 – 18)
= 5 x 0 = 0

Question 10.
Find the sum of the natural numbers between 101 and 999 which are divisible by both 2 and 5.
Solution:
Natural numbers between 101 and 999 which are divisible by 2 and 5 both are 110, 120, 130, …, 990
Here a = 110 and d= 10, l = 990
Now, l = a_n = a + (n – 1 )d
990 = 110 + (n – 1) x 10
⇒ 990 – 110 = 10(n – 1) ⇒ 10(n – 1) = 880
⇒ n – 1 = \(\frac { 880 }{ 10 }\) ⇒ n – 1 = 88
∴ n = 88 + 1 = 89
Now, S₈₉ = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
= \(\frac { 89 }{ 2 }\) [2 x 110 + (89 – 1) x 10]
= \(\frac { 89 }{ 2 }\) [220 + 88 x 10] = \(\frac { 88 }{ 2 }\) [220 + 880]
= \(\frac { 89 }{ 2 }\) x 1100 = 48950

Question 11.
Find the sum of ail two digits numbers greatest than 50 which when divided by 7, leave a remainder of 4.
Solution:
2-digit number greater than 50 which are divisible by 7, leaves a remainder of 4 are: 53, 60, 67, 74, 81, 88, 95
Here a = 53, d = 7, l = 95
∴ a_n(l) = a + (n – 1)d
95 = 53 + (n – 1) x 7 ⇒ 95 – 53 = 7(n – 1)
⇒ 7(n – 1) = 42 ⇒ n – 1 = \(\frac { 42 }{ 7 }\) = 6
∴ n = 6 + 1 = 7
Now, S_n = \(\frac { n }{ 2 }\)[2a + (n – 1 )d]
= \(\frac{7}{2}[2 \times 53+(7-1) \times 7]\)
= \(\frac { 7 }{ 2 }\)[106 + 6 × 7]
= \(\frac { 1 }{ 2 }\)

Question 12.
Find the sum of the integers between 100 and 200 that are (i) divisible by 9 (ii) not divisible by 9.
Solution:
Integers between 100 and 200
(i) Which are divisible by 9 are
108, 117, 126, …, 198
Here, a = 108, d = 9, l = 198
∴ a_n(l) = a + (n – 1 )d
⇒ 198 = 108 + (n – 1) x 9
⇒ 198 – 108 = 9(n – 1)
⇒ 90 = 9 (n – 1)
⇒ n – 1 = \(\frac { 90 }{ 9 }\) = 10
∴ n = 10 + 1 = 11
Now, S₁₁ = \(\frac { n }{ 2 }\) [2a + (n – 1)d]
= \(\frac { 11 }{ 2 }\)[2 x 108 + (11 – 1) x 9]
= \(\frac { 11 }{ 2 }\)[216 + 10 x 9] = \(\frac { 11 }{ 2 }\)[216 + 90]
= \(\frac { 11 }{ 2 }\)[306] = 1683

(ii) Now, sum of integers from 101 to 199
Here, a = 101, d = 1, l = 199, n = 99
S_n = \(\frac { n }{ 2 }\) [2a + (n – 1)d]
= \(\frac { 99 }{ 2 }\)[2 x 101 +(99 – 1) x 1]
= \(\frac { 99 }{ 2 }\) [202 + 98] = \(\frac { 99 }{ 2 }\) x 300 = 14850
∴ Sum of integers which are not divisible by 9 = 14850 – 1683 = 13167

Question 13.
Find the middle term of the sequence formed by all the numbers between 9 and 95 which leave a remainder 1 when divided by 3. Also find the sum of numbers on both the sides of the middle term separately.
Solution:
Numbers between 9 and 95 when divided
by 3, leaves remainder 1
10, 13, 16, 19, …, 94
Here, a = 10, d = 3, l = 94
a_n(l) = a + (n – 1)d
94 = 10 + (n – 1) x 3 ⇒ 94 – 10 = 3 (n – 1)
⇒ 84 = 3 (n – 1) ⇒ n – 1 = \(\frac { 84 }{ 3 }\) = 28
∴ n = 28 + 1 = 29
Now middle term = \(\frac { 29+1 }{ 2 }\)th = 15th term Sum of first 14 terms
= \(\frac { n }{ 2 }\)[2 a + (n – 1 )d]
= \(\frac { n }{ 2 }\)[2a + (14 – 1)d]
= \(\frac { 14 }{ 2 }\)[2 x 10 + 13 x 3]
= 7[20 + 39] = 59 x 2 = 413
Middle term = 10 + 15 x 3 = 10 + 45 = 55
After middle term, number of terms = 14
Whose first term (a) = 55, d = 3 and n = 14 Similarly,
S₁₄ = \(\frac { 19 }{ 2 }\)[2 x 55 + (14 – 1) x 3]
= \(\frac { 19 }{ 2 }\)[110 + 13 x 13] = 7[110 + 39]
= 7 x 149 = 1043

Question 14.
If Sn denotes the sum of first n terms of an A.P., prove that S₁₂ = 3(S₈ – S₄).
Solution:
S_n = Sum of first n terms of an A.P.
S_n = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
To prove : S₁₂ = 3(S₈ – S₄)
RHS = 3[S₈ – S₄]
= 3[ \(\frac { 8 }{ 2 }\)[2a + (8 – 1)d] – \(\frac { 4 }{ 2 }\)[2a + (4 – 1)d] ]
= 3[ \(\frac { 8 }{ 2 }\) (2a + 7J) – \(\frac { 4 }{ 2 }\)(2a + 3d) ]
= 3 [4(2 a + 7d) – 2(2 a + 3d)]
= 3[8a + 28 d – 4a – 6d]
= 3[4a + 22d] = 12a + 66d
= 6[2 a+ 11 d]
= \(\frac { 12 }{ 2 }\) [2a + (12 – 1)d] = S₁₂ = LHS

Question 15.
Show that the sum of first n even natural numbers is equal to (1 + \(\frac { 1 }{ n }\)) times the sum of the first n odd natural numbers.
Solution:
Sum of first n even natural number = (1 + \(\frac { 1 }{ n }\))
(Sum of first n odd natural numbers) Sum of first n even natural numbers (2, 4, 6, …, 2n)
= \(\frac { n }{ 2 }\) [2a + (n – 1 )d]
= \(\frac { n }{ 2 }\) [2 x 2 + (n- 1) x 2]
= \(\frac { n }{ 2 }\) [4 + 2n – 2]
= \(\frac { n }{ 2 }\) [2 + 2n]
= n( 1 + n)
and sum of n odd natural numbers (1, 3, 5, …., (2n – 1)
= \(\frac { n }{ 2 }\) [2x l + (n – 1)2]
= \(\frac { n }{ 2 }\) [2 + 2n – 2]
= \(\frac { n }{ 2 }\)[2n] = n²
Now, n² x (1 + \(\frac { 1 }{ n }\)) = n²(\(\frac { n+1 }{ n }\))
= n( 1 + n)

Question 16.
The sum of n terms of an A.P. whose first term is 5 and common difference is 36 is equal to the sum of In terms of another A.P. whose first term is 36 and common difference is 5. Find n.
Solution:
First term (a₁) = 5
Common difference (d₁) = 36
S_n = Sum of 2n terms of another AP whose
first term (a₂) = 36 and common difference
(d₂) = 5
In first AP
S_n = \(\frac { n }{ 2 }\)[2a + (n- 1 )d]
= \(\frac { n }{ 2 }\)[2 x 5 + (n – 1) x 36]
= \(\frac { n }{ 2 }\) [10 + 36n – 36] = \(\frac { n }{ 2 }\)[36n – 26]
= n(18n – 13)
In second AP,
S_2n= \(\frac { n }{ 2 }\) [2 x 36 + (2n – 1)5]
= n[72 + 10n – 5]
= n[67 + 10n]
∵ s_n = s_2n
n(18n – 13) = n(67 + 10n)
⇒ 18n – 10n = 67 + 13 ⇒ 8n = 80
n = \(\frac { 80 }{ 8 }\) = 10
∴ n = 10

Question 17.
If the sum of first ten terms of an A.P. is 4 times the sum of the first five terms, then find the ratio of the first term of the common difference.
Solution:
Sum of first 10 terms of an AP = 4 x Sum of first 5 terms
Let a be the first term and d be the common difference, then
S_n = \(\frac { n }{ 2 }\)[2 a + (n – 1 )d]
S₁₀ = \(\frac { 10 }{ 2 }\) [2a + (10 – 1)d = 5(2a + 9d)
and S₅ = \(\frac { 5 }{ 2 }\) [2a + (5 – 1 )d] = \(\frac { 5 }{ 2 }\) [2a + 4d]
According to the condition,
5(2 a + 9d) = 4 x \(\frac { 5 }{ 2 }\) (2a + 4d)
10a + 45d = 20a + 40d
45d – 40d = 20a – 10a
⇒ 10a = 5d ⇒ \(\frac { a }{ d }\) = \(\frac { 5 }{ 10 }\) = \(\frac { 1 }{ 2 }\)
∴ Ratio in first term to the common difference = 1 : 2

Question 18.
In a nursery, 37 plants have been arranged in the first row, 35 in the second, 33 in the third and so on. If there are 5 plants in the last row, how many plants are there in the nursery.
Solution:
In the first row, plants are = 37
In second row = 35
In third row = 33
and in the last row = 5
Here, a = 37, d = 35 – 37 = – 2
a_n = a + (n – 1)d
5 = 37 + (n – 1)(- 2)
5 – 37 = – 2(n – 1)
⇒ – 32 = – 2(n – 1)
n – 1 = \(\frac { -32 }{ -2 }\) = 16
∴ n = 16 + 1 = 17
∴ Number of rows = 17
Now, total plants (S_n) = \(\frac { n }{ 2 }\) [2a + (n – 1 )d]
= \(\frac { 17 }{ 2 }\)[2 x 37 + (17 – 1)(- 2)]
= \(\frac { 17 }{ 2 }\) [74 + 16 x (- 2)]
= \(\frac { 17 }{ 2 }\)[74 – 32] = \(\frac { 17 }{ 2 }\) x 42 = 357
∴ Total plants = 357

Question 19.
200 logs are stacked in the following manner : 20 legs in the bottom row, 19 in the next, above it, 18 in the row above it and so on. In how many rows are the 200 logs placed and how many logs are there in the top row?
Solution:
Total logs = 200
In the bottom row, number of logs = 20
and next row above it = 19
Next row above it = 18
and so on
So, AP is 20, 19, 18, 17, …
and S_n = 200
Let the top row be the nth row
∴ a_n = a + (n – 1 )d
and S_n = 200
∴ s_n = \(\frac { n }{ 2 }\)[2 a + (n – 1 )d]
200 = \(\frac { n }{ 2 }\) [2 x 20 + (n – 1)(- 1)]
⇒ 400 = n(40 – n + 1)
⇒ 400 = 41n – n²
⇒ n² – 41n + 400 = 0
⇒ n² – 25n – 16n + 400 = 0
⇒ n(n – 25) – 16(n – 25) = 0
⇒ (n – 25) (n – 16) = 0
Either n – 25 = 0, then n = 25
or n – 16 = 0, then n = 16
But n = 25 is not possible
As number of logs in the first row = 20
and d = – 1
∴ Number by rows = 16
and number of log in 16th row
= a + (n – 1)d = 20 + (16 – 1) (- 1)
= 20 + 15(- 1) = 20 – 15 = 5 logs

Question 20.
A sum of ₹ 1890 is to be used to give seven cash prizes. If each prize is ₹ 50 less than the preceding prize, find the value of each prize?
Solution:
Total amount = ₹ 1890
and number of cash prizes = 7
and each prize is 750 less than the proceeding
prize.
Let first prize = ₹ a
Then second prize a – 50
Third prize = a – 100 and so on
Here, first term = a and d = – 50, S_n = 1890 n = 7
s_n = \(\frac { n }{ 2 }\) [2a + (n – 1)d]
1890 = \(\frac { n }{ 2 }\)[2a + (7 – 1)(- 50)]
\(\frac { 2×1890 }{ 7 }\) = 2a – 300
540 + 300 = 2a ⇒ 2a = 840
a = \(\frac { 840 }{ 2 }\) = 420
Prizes are ₹ 420, ₹ 370, ₹ 320, ₹ 270, ₹ 220, ₹ 170 and ₹ 120