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

Practicing OP Malhotra Class 10 ICSE Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(a) is the ultimate need for students who intend to score good marks in examinations.

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

Question 1.
Which of the following situations does the list of numbers involved make an arithmetic progression and why?
(i) the amount of air present in a balloon when \(\frac { 1 }{ 5 }\)th of the air remaining in the balloon is removed by a vacuum pump at a time.
(ii) the cost of digging a well after every metre of digging, if it costs ₹ 125 for the first metre and increases by ₹ 65 for every subsequent metre.
Solution:
(i) Yes.

(ii) Yes, it is an arithmetic progression as cost of digging in the beginning is ₹ 125 for the first metre and then increasing ₹ 65 for every subsequent metre.

Question 2.
Write the first five terms of an AP. in which
(i) a = 10, d = 7
(ii) a = 55, d = – 4
(iii) a = – 1, d = \(\frac { 1 }{ 2 }\)
(iv) a = – 3.25, d = – 0.25
Solution:
(i) a = 10, d = 7
∴ 10, 17, 24, 31, 38

(ii) a = 55, d = -4
∴AP = 55, 51, 47, 43, 39

(iii) a = – 1, d = \(\frac { 1 }{ 2 }\)
∴ AP = – 1, – \(\frac { 1 }{ 2 }\), 0, \(\frac { 1 }{ 2 }\), 1

(iv) a = – 3.25, d = – 0.25
∴ AP = – 3.25, – 3.50, – 3.75, – 4.00, – 4.25

Question 3.
Choose the correct answer in the following:
(i) the list of numbers -12, -9, -6, -3, 0,3 is
(a) not an AP
(b) an AP with d = – 3
(c) an AP with d = 0
(d) an AP with d- 3

(ii) In an AP if a = 3, d = 0 and n = 7, then a_n will be
(a) 4
(b) 1
(c) 3
(d) 2
Solution:
(i) List of numbers : -12, -9, -6, -3, 0, 3
Here a = -12, d = -9 – (-12) = – 9 + 12 = 3
a = – 12, d = 3

(ii) In an AP, a = 3, d = 0, n = 7, then
a_n = a + (n – 1 )d = 3 + (7 – 1) + 0
= 3 + 0 = 3

Question 4.
What is the:
(i) 25th term of the A.P.: – 5, – \(\frac { 5 }{ 2 }\), 0, \(\frac { 5 }{ 2 }\), …
(ii) 30th term of the A.P.: 27, 22, 17, …
(iii) 10th term of the A.P.: -0.1, -0.2, -0.3,…
Solution:
(i) AP = – 5, – \(\frac { 5 }{ 2 }\), 0, \(\frac { 5 }{ 2 }\), …
Here, a = -5, d = – \(\frac { 5 }{ 2 }\) – (- 5)
∴ a₂₅ = a + (n – 1 )d = – 5 + (25 – 1) x \(\frac { 5 }{ 2 }\)
= – 5 + 24 x \(\frac { 5 }{ 2 }\) = – 5 + 60 = 55

(ii) AP is 27, 22, 17, …
Here, a = 27, d= (22 – 27) = – 5
∴ a₃₀ = a + (n – 1 )d
= 27 + (30 – 1) x (- 5) = 27 + 29(- 5)
= 27 – 145 = – 118

(iii) AP = -0.1, -0.2, -0.3, …
Here, a = -0.1
d = – 0.2 – (- 0.1) = – 0.2+ 0.1 = – 0.1
∴ a₁₀ = a + (n – 1)d = – 0.1 +(10 – 1) (- 0.1)
= – 0.1 + 9 x (- 0.1)
= – 0.1 – 0.9 = – 1

Question 5.
If k, 2k – 1 and 2k + 1 are the three consecutive terms of an AP, then find the value of k.
Solution:
k, 2k – 1 and 2k + 1 are the three consecutive terms of an AP, then
a = k, d = 2k – 1 – k = k – 1 … (i)
and d = 2k + 1 – (2k – 1) = 2 k + 1 – 2k + 1 = 2 …(ii)
From (i) and (ii),
∴ k – 1 = 2 ⇒ k = 2 + 1 = 3
∴ k = 3

Question 6.
What is the common difference of the
A.P. \(\frac{1}{3 q}, \frac{1-6 q}{3 q}, \frac{1-12 q}{3 q}\).
Solution:
AP is \(\frac{1}{3 q}, \frac{1-6 q}{3 q}, \frac{1-12 q}{3 q}\)
d = \(\frac{1-6 q}{3 q}-\frac{1}{3 q}=\frac{1-6 q-1}{3 q}=\frac{-6 q}{3 q}\)
= – 2
∴ Common difference = – 2

Question 7.
Which term of the A.P., 5, 13, 21, … is 181?
Solution:
Which term of AP 5, 13, 21, … is 181
Let it be nth term, then
In AP a = 5, d = 13 – 5 = 8
a_n = a + (n – 1 )d
181 =5 + (n- 1) x 8 ⇒ 181 = 5 + 8n – 8
⇒ 8n = 181 – 5 + 8 = 184 ⇒ n = \(\frac { 1 }{ 2 }\) = 23
∴ 181 is 23rd term.

Question 8.
Which term of the A.P. 3, 10, 17,… will be 84 more than its 13th term?
Solution:
AP is 3, 10, 17,…
Here, a = 3, d = 10 – 3 = 7
Let nth term is 84 more than its 13th term
a₁₃ = a + (n – 1)d = 3 + (13 – 1) x 3
= 3 + 12 x 7 = 87
a_n = 3 + (n – 1) x 7 = 3 + 7n – 7 = 7n – 4
Now, 7n – 4 = 84 + 87 = 171
7n = 171 + 4 = 175
n = \(\frac { 175 }{ 7 }\)
∴ 25th is the required term.

Question 9.
Find the AP if the 6th term of the A.P. is 19 and the 16th term is 15 more than the 11th term.
Solution:
In an AP
6th term = 19 and
16th term = 15 + 11th term
Let a be the first term and d be the common difference
∴ a₆ = a + (n – 1 )d ⇒ 19 = a + 5d … (i)
⇒ a + 5d = 19
Similarly,
a₁₁ = a + 10d and a₁₆ = a + 15d
Now, a + 15d = a + 10d + 15
15d – 10d = 15 ⇒ 5d = 15
⇒ d = \(\frac { 15 }{ 5 }\) = 3
From (i)
a + 5 x 3 = 19 ⇒ a + 15 = 19
⇒ a = 19 – 15 = 4
or = 4, d = 3
∴ AP is 4, 7, 10, 13, 16, …

Question 10.
The sum of the 2nd and 7th terms of an A.P. is 30. If its 15th term is 1 less than twice its 8th term, find the AP.
Solution:
2nd term + 7th term = 30
15th term = 2 x 8th term – 1
Let a be the first term and d be the common
difference, then
a₂ = a + (n – 1 )d = a + (2 – 1 )d = a + d
Similarly, a₇ = a + (7 – 1)d ⇒ a + 6d
a₁₅ = a + (15 – 1)d ⇒ a + 14d and
a₈ = a + (8 – 1 )d ⇒ a + 7d
Now, a₂ + a₇ = 30 ⇒ a + d + a + 6d = 30
⇒ 2a + 7d = 30 … (i)
and a + 14d = 2 x (a + 7d) – 1
a + 14d = 2a + 14d – 1
2a – a – 1 ⇒ a = 1
From (i), 2 x 1 + 7d= 30 ⇒ 2 + 7d = 30
⇒ 7d = 30 – 2 = 28
⇒ d = \(\frac { 28 }{ 7 }\) ⇒ d = 4
∴ d = 4, a = 1
∴ AP = 1, 5, 9, 13, 17, …

Question 11.
The fourth term of an A.P. is 11. The sum of the fifth and seventh terms of the A.P. is 34, find its common difference.
Solution:
In an AP
4th term (a₄) = 11
5th term + 7th term = 34
Let a be the first term and d be the common
difference, then
a₄ – a + (n – 1 )d = a + (4 – 1)d
⇒ a + 3d = 11 … (1)
Similarly,
a₇ = a + (7 – 1 )d ⇒ a + 6d
We know that, a₄ + a₇ = 34
∴ a + Ad + a + 6d = 34
⇒ 2a + 10d = 34
a + 5d = 17 … (ii)
Subtracting (ii) from (i),
= (a + 3d) – (a + 5d) = 11 – 17
= a + 3d – a – 5d = – 6
– 2d = – 6
2d = 6 ⇒ d = \(\frac { 6 }{ 2 }\) = 3
∴ Common difference = 3

Question 12.
Find the middle term of the A.P. 213,205, 197, …, 37.
Solution:
AP = 213, 205, 197, …, 37
Let 37 be the nth term
Now, a = 213 and d = 205 – 213 = – 8
∴ a_n = a + (n – 1 )d
⇒ 37 = 213 + (n – 1) (- 8)
⇒ 37 = 213 – 8n + 8
8n = 213 + 8 – 37 = 221 – 37 = 184
n = \(\frac { 184 }{ 8 }\) = 23
∴ There are 23 term in the AP 23 + 1
∴ Middle term = \(\frac { 23+1 }{ 2 }\) = 12th term
Nw, a₁₂ = a + 11 d
= 213 + 11 (- 8) = 213 – 88 = 125

Question 13.
If the 3rd and 9th terms of an A.P. are 4 and – 8 respectively, then which term of this A.P. is zero?
Solution:
In an AP
3rd term = 4
9th term = – 8
Which term is 0
Let a be the first term and d be the common difference, then
a₃ = a + (n – 1 )d
⇒ 4 = a + (3 – 1)d
⇒ a + 2d = 4 … (i)
Similarly, a + ad = – 8 … (ii)
Subtracting (i) from (ii),
⇒ (a + 8d) – (a + 2d) = – 8 – 4
6d = – 12 ⇒ d = \(\frac { -12 }{ 6 }\) = – 2
and a + 2(- 2) = 4 ⇒ a – 4 = 4
⇒ a = 4 + 4 = 8
∴ a = 8, d = – 2
Let a_n be equal to 0, then
a + (n – 1 )d = 0
⇒ 8 + (n – 1) (- 2) = 0
⇒ + 8 – 2n + 2 = 0
⇒ 10 – 2n = 0
⇒ 2n = 10
n = \(\frac { 10 }{ 2 }\) = 5
∴ 5th term is zero.

Question 14.
The 8th term of an A.P. is zero. Prove that its 38th term is three times its 18th term.
Solution:
In an AP
8th term (a₈) = 0
To prove that 38th term = 3 x 18th term
Let a be the first term and d be the common difference, then
a₈ = a + (n – 1)d
⇒ a + (8 – 1)d = 0
⇒ a + 7d = 0
⇒ a = – 7d … (i)
Similarly,
a₁₈ = a + 17d
a₃₈ = a +31d
Now, a + 17d = – 7d + 17d = 10d
and a + 37d = – 7d + 37d = 30d
∴ 30d = 3 x 10d = 30d
Hence, a₃₈ = 3 x a₁₈

Question 15.
Which term of the A.P. 120, 116, 112,… is its first negative term?
Solution:
AP is 120, 116, 112,…
Which first term of this AP will be negative
Here, a = 120, d = 116 – 120 = – 4
Let nth term of the given AP be the negative term a_n or T_n < 0
⇒ a + (n – 1)d < 0
⇒ 120 + (n – 1) (- 4) < 0
⇒ 120 – 4n + 4 < 0
⇒ 124 – 4n < 0 ⇒ 4n > 124
⇒ n > \(\frac { 124 }{ 4 }\) = 31
∴ n = 32 be the first negative term.
⇒ 32nd term is the first negative term

Question 16.
How many 3-digit natural numbers are divisible by 7?
Solution:
3-digit terms are 100, 101, 102, …, 999
and terms which are divisible by 7 will be 105, 112, 119, …, 994
Here, a = 105, d = 7
Last term (a_n) = 994
∴ a_n = a + (n – 1 )d
994 = 105 + (n – 1) x 7
= 105 + 7K – 7 = 7n + 98
∴ 7n = 994 – 98 = 896
n = \(\frac { 896 }{ 7 }\) = 128
∴ Number of 3-digit term divisible by 7 = 128

Question 17.
Two APs have the same common difference. The first term of one of these is – 1 and that of the other is – 8. What is the difference between their 4th terms?
Solution:
Let d be the common difference of the two AP series ax and a2 are the first term of the two AP’s respectively.
In first AP, a₁ = 1, a₂ = – 8
Difference between their 4th terms = (a₁ + 3d) – (a₂ + 3d) = – 1 – (- 8)
a₁ + 3d – a₂ – 3d = – 1 + 8 ⇒ a₁ – a₂ = 7
or a₂ + 3d – a₁ – 3d = – 8 – (- 1)
a₂ – a₁ = – 8 + 1 = – 7
∴ Difference = 7 or – 7

Question 18.
If an AP, ratio of the 4th and 9th terms is 1 : 3, find the ratio of 12 term and 5th term?
Solution:
In an AP
Ratio between 4th term and 9th term = 1 : 3
Let a be the first term and d be the common difference, then
4th term (a₄) = a + (n – 1 )d= a + (4 – 1 )d = a + 3d
Similarly 9th term = a + 8d
Now,
∴ \(\frac{a+3 d}{a+8 d}=\frac{1}{3}\)
3a + 9d = a + 8d
⇒ 3a – a = 8d – 9d
⇒ 2a = – d
⇒ d = – 2a
Now, a₁₂ = a + (12 – 1 )d = a + 11d
and a₅ = a + (5 – 1 )d = a + 4d
∴ \(\frac{a+11 d}{a+4 d}=\frac{a+(-2 a) \times 11}{a+4(-2 a)}\)
\(\frac{a-22 a}{a-8 a}=\frac{-21 a}{-7 a}\)
= \(\frac { 3 }{ 1 }\) = 3 : 1

Question 19.
Find the 7th term from the end of the
A.P.7,10, 13 184.
Solution:
7th term from the end of AP
7, 10, 13, …, 184
Here, a = 7, d = 10 – 7 = 3,
T_n(l) = 184
T_n = a + (n – 1)d
⇒ 7 +(n – 1) x 3
⇒ 7 + 3n – 3 = 4 + 3n
∴ 4 + 3M = 184
⇒ 3n = 184 – 4 = 180
n = \(\frac { 180 }{ 3 }\) = 60
7th term from the last will be 60 – (7 – 1) = 60 – 6 = 54th from the beginning.
∴ T₅₄ = a + 53d
= 7 + 53 x 3 = 7 + 159 = 166

Question 20.
The four angles of a quadrilateral form an A.P. If the sum of the first three angles is twice the fourth angle, then find all the angles.
Solution:
Four angles of a quadrilateral are in AP
Sum of the first 3 angles = 2 x 4th angle
Adding 4th angle to both sides
Sum of 4 angles = 2 x 4th angle + 4th angle = 3 x 4th angle
But sum of 4 angles of a quadrilateral = 360
∴ 3 x 4th angle = 360°
⇒ 4th angle = \(\frac { 360° }{ 3 }\) = 120°
⇒ 4th angle = 120°
Let a be the first term (angle) and d be the common difference.
∴ a + 3d= 120° … (i)
and a + a + d+a + 2d = 360° – 120° = 240
3a + 3d = 240 … (ii)
Subtracting 2a = 120 ⇒ a = \(\frac { 120° }{ 2 }\) = 60°
But a + 3d = 120
⇒ 60 + 3d = 120 ⇒ 3d = 120 – 60 = 60
d = \(\frac { 60° }{ 3 }\) = 20°
∴ Angles are 60°, 80°, 100°, 120°

Question 21.
The sum of three numbers in A.P. is – 3 and their product is 8. Find the numbers.
Solution:
Sum of 3 numbers in AP = – 3
and product = 8
Let three numbers in AP be
a – d, a, a + d
∴ a – d + a + a + d = – 3
⇒ 3a = – 3 ⇒ a = \(\frac { -3 }{ 3 }\) = – 1
and (a – d)a(a + d) = 8
⇒ (a² – d²)a = 8
[(- 1 )² – d²)](-1) = 8
1 – d² = – 8
d² = 1 + 8
d² = 9
d₂ = (±3)²
∴ d = + 3
∴ Numbers be – 1 – 3, – 1, – 1 + 3
= – 4, – 1, 2
or – 1 + 3, – 1, – 1 – 3
= 2, – 1, – 4

Question 22.
Ram prasad saved ₹ 10 in the first week of the year and then increased his weekly savings by ₹ 2.75. If in the nth week, his savings become ₹ 59.50, find n.
Solution:
Ram Prasad’s savings in first week = ₹ 10
Then his weekly saving is increasing = ₹ 2.75
∴ a = ₹ 10 and d = ₹ 2.75
nth week saving = ₹ 59.50
a_nn = a + {n – 1 )d
59.50 = 10 + (n – 1)(2.75)
59.50 – 10.00 = (n – 1)(2.75)
49.50 = (n – 1)(2.75)
∴ n – 1 = \(\frac { 49.50 }{ 2.75 }\)
∴ n = 18 + 1 = 19

Question 23.
For an A.P. show that T_p + T_p+2q = 2T_p+q.
Solution:
For an AP
T_p + T_p+2q = 2T_p+q.
Let a be the first term and d be the common difference, then
T_p = a + (p – 1)d,
T_p+2q = a + (p + 2q – 1)d
T_p+q = a + (p + q – 1)d
Now, LHS = T_p + T_p+2q = a + (p – 1)d + a + (p + 2q – 1 )d
= a + dp – d + a + pd + 2qd – d
= 2a + 2dp + 2qd – 2d
= 2(a + dp + qd – d)
= 2[a + (p + q – 1)d]
= 2 x T_p+q
= RHS