Class 10 SELINA Solutions Maths Chapter 10 - Arithmetic Progression
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When it comes to ICSE Class 10 Math, the chapter Arithmetic Progressions holds a lot of importance. It requires students to use basic mathematical operations, identify patterns, calculate terms and sums, explore number patterns, and develop sequencing skills. Students must simplify expressions and use their mathematical reasoning, problem-solving abilities, and pattern recognition skills to analyze and manipulate sequences of numbers with a constant difference. To solve such questions, students need a solid understanding of mathematical concepts and the application of appropriate techniques.
Selina Solutions offer comprehensive and detailed solutions to the problems in the Arithmetic Progressions chapter. They cover all the topics and concepts from the ICSE Class 10 Mathematics syllabus. Explanations and step-by-step answers ensure that the students understand the underlying principles and techniques required to solve arithmetic progression problems.
Students get exposed to different problems, from simple to complex arithmetic progressions. This allows them to practice various fractions, decimals, percentages, ratios, proportions, and more. Thus, the solutions offer a range of examples which helps students build proficiency, self-assess their progress and recognise concepts that require further practice or clarification.
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Arithmetic Progression Exercise Ex. 10(A)
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
The given A.P. is 1, 4, 7, 10, ……….
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
For a given A.P.,
Number of terms, n = 60
First term, a = 7
Last term, l = 125
⇒ t60 = 125
⇒ a + 59d = 125
⇒ 7 + 59d = 125
⇒ 59d = 118
⇒ d = 2
Hence, t31 = a + 30d = 7 + 30(2) = 7 + 60 = 67
Solution 19
Let 'a' be the first term and 'd' be the common difference of the given A.P.
t4 + t8 = 24 (given)
⇒ (a + 3d) + (a + 7d) = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 ….(i)
And,
t6 + t10 = 34 (given)
⇒ (a + 5d) + (a + 9d) = 34
⇒ 2a + 14d = 34
⇒ a + 7d = 17 ….(ii)
Subtracting (i) from (ii), we get
2d = 5
Solution 20
Let 'a' be the first term and 'd' be the common difference of the given A.P.
Now, t3 = 5 (given)
⇒ a + 2d = 5 ….(i)
And,
t7 = 9 (given)
⇒ a + 6d = 9 ….(ii)
Subtracting (i) from (ii), we get
4d = 4
⇒ d = 1
⇒ a + 2(1) = 5
⇒ a = 3
Hence, 17th term = t17 = a + 16d = 3 + 16(1) = 19
Arithmetic Progression Exercise Ex. 10(B)
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Arithmetic Progression Exercise Ex. 10(C)
Solution 1
Solution 2
Solution 3
Solution 4(i)
Solution 4(ii)
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Arithmetic Progression Exercise Ex. 10(D)
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
We know that,
Sum of n terms of an A.P = 
Let the first term be 2x and the last term be 3x.
∴ Sum of 5 terms of an A.P = 
First term = 2x =2 × 1 = 2 and the last term = 3x = 3 × 1 = 3
nth term of an A.P. is given by
tn = a + (n - 1)d
⇒ a5 = 2 + (5 - 1)d
⇒ 3 = 2 + 4d
⇒ 1 = 4d
⇒ d = 
= 0.25
Therefore the five numbers in an A.P are 2, 2.25, 2.50, 2.75 and 3.
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Arithmetic Progression Exercise Ex. 10(E)
Solution 1
Solution 2
Solution 3
Solution 4
Since the production increases uniformly by a fixed number every year, he sequence formed by the production in different years is an A.P.
Let the production in the first year = a
Common difference = Number of units by which the production increases every year = d
Solution 5
Total amount of loan = Rs. 1,18,00
First instalment = a = Rs. 1000
Increase in instalment every month = d= Rs. 100
30th instalment = t30
= a + 29d
= 1000 + 29 × 100
= 1000 + 2900
= Rs. 3900
Now, amount paid in 30 instalment = S30
= 15 × 4900
= Rs. 73, 500
∴ Amount of loan to be paid after the 30th instalment
= Rs. (1,18,000 - 73500)
= Rs. 44,500
Arithmetic Progression Exercise Ex. 10(F)
Solution 1
Let 'a' be the first term and 'd' be the common difference of the given A.P.
Now, t6 = 16 (given)
⇒ a + 5d = 16 ….(i)
And,
t14 = 32 (given)
⇒ a + 13d = 32 ….(ii)
Subtracting (i) from (ii), we get
8d = 16
⇒ d = 2
⇒ a + 5(2) = 16
⇒ a = 6
Hence, 36th term = t36 = a + 35d = 6 + 35(2) = 76
Solution 2
Solution 3
For a given A.P.,
Number of terms, n = 50
3rd term, t3 = 12
⇒ a + 2d = 12 ….(i)
Last term, l = 106
⇒ t50 = 106
⇒ a + 49d = 106 ….(ii)
Subtracting (i) from (ii), we get
47d = 94
⇒ d = 2
⇒ a + 2(2) = 12
⇒ a = 8
Hence, t29 = a + 28d = 8 + 28(2) = 8 + 56 = 64
Solution 4
Solution 5
Here,
First term, a = 4
Common difference, d = 6 - 4 = 2
n = 10
Solution 6
Here,
First term, a = 3
Last term, l = 57
n = 20
Solution 7
Here, we find that
15 - 18 = 12 - 15 = -3
Thus, the given series is an A.P. with first term 18 and common difference -3.
Let the number of term to be added be 'n'.
⇒ 90 = n[36 - 3n + 3]
⇒ 90 = n[39 - 3n]
⇒ 90 = 3n[13 - n]
⇒ 30 = 13n - n2
⇒ n2 - 13n + 30 = 0
⇒ n2 - 10n - 3n + 30 = 0
⇒ n(n - 10) - 3(n - 10) = 0
⇒ (n - 10)(n - 3) = 0
⇒ n - 10 = 0 or n - 3 = 0
⇒ n = 10 or n = 3
Thus, required number of term to be added is 3 or 10.
Solution 8
tn = 8 - 5n
Replacing n by (n + 1), we get
tn+1 = 8 - 5(n + 1) = 8 - 5n - 5 = 3 - 5n
Now,
tn+1 - tn = (3 - 5n) - (8 - 5n) = -5
Since, (tn+1 - tn) is independent of n and is therefore a constant.
Hence, the given sequence is an A.P.
Solution 9
The given sequence is 1, -1, -3, …..
Now,
1 - 3 = -1 - 1 = -3 - (-1) = -2
Hence, the given sequence is an A.P. with first term a = 3 and common difference d = -2.
The general term (nth term) of an A.P. is given by
tn = a + (n - 1)d
= 3 + (n - 1)(-2)
= 3 - 2n + 2
= 5 - 2n
Hence, 23rd term = t23 = 5 - 2(23) = 5 - 46 = -41
Solution 10
The given sequence is 3, 8, 13, …..
Now,
8 - 3 = 13 - 8 = 5
Hence, the given sequence is an A.P. with first term a = 3 and common difference d = 5.
Let the nth term of the given A.P. be 78.
⇒ 78 = 3 + (n - 1)(5)
⇒ 75 = 5n - 5
⇒ 5n = 80
⇒ n = 16
Thus, the 16th term of the given sequence is 78.
Solution 11
The given sequence is 11, 8, 5, 2, …..
Now,
8 - 11 = 5 - 8 = 2 - 5 = -3
Hence, the given sequence is an A.P. with first term a = 11 and common difference d = -3.
The general term of an A.P. is given by
tn = a + (n - 1)d
⇒ -150 = 11 + (n - 1)(-5)
⇒ -161 = -5n + 5
⇒ 5n = 166
The number of terms cannot be a fraction.
So, clearly, -150 is not a term of the given sequence.
Solution 12
Solution 13
Solution 14
Solution 15
Let 'a' be the first term and 'd' be the common difference of the given A.P.
Given,
S14 = 1050
⇒ 7[2a + 13d] = 1050
⇒ 2a + 13d = 150
⇒ a + 6.5d = 75 ….(i)
And, t14 = 140
⇒ a + 13d = 140 ….(ii)
Subtracting (i) from (ii), we get
6.5d = 65
⇒ d = 10
⇒ a + 13(10) = 140
⇒ a = 10
Thus, 20th term = t20 = 10 + 19d = 10 + 19(10) = 200
Solution 16
nth term of an A.P. is given by tn= a + (n - 1) d.
⇒ t25 = a + (25 - 1)d = a + 24d and
t9 = a + (9 - 1)d = a + 8d
According to the condition in the question, we get
t25 = t9 + 16
⇒ a + 24d = a + 8d + 16
⇒ 16d = 16
⇒ d = 1
Solution 17
Let a and d be the first term and common difference respectively.
⇒(m + n)th term = a + (m + n - 1)d …. (i) and
(m - n)th term = a + (m - n - 1)d …. (ii)
From (i) + (ii), we get
(m + n)th term + (m - n)th term
= a + (m + n - 1)d + a + (m - n - 1)d
= a + md + nd - d + a + md - nd - d
= 2a + 2md - 2d
= 2a + (m - 1)2d
= 2[ a + (m - 1)d]
= 2 × mthterm
Hence proved.
Solution 18
In the first A.P. 58, 60, 62,....
a = 58 and d = 2
tn = a + (n - 1)d
⇒ tn = 58 + (n - 1)2 …. (i)
In the first A.P. -2, 5, 12, ….
a = -2 and d = 7
tn = a + (n - 1)d
⇒ tn= -2 + (n - 1)7 …. (ii)
Given that the nth term of first A.P is equal to the nth term of the second A.P.
⇒58 + (n - 1)2 = -2 + (n - 1)7 … from (i) and (ii)
⇒58 + 2n - 2 = -2 + 7n - 7
⇒ 65 = 5n
⇒ n = 15
Solution 19
Here a = 105 and d = 101 - 105 = -4
Let an be the first negative term.
⇒ an < 0
⇒ a + (n - 1)d < 0
⇒ 105 + (n - 1)(-4)<0
⇒ 105 - 4n + 4 <0
⇒ 109 - 4n < 0
⇒ 109 <4n
⇒ 27.25 < n
The value of n = 28.
Therefore 28th term is the first negative term of the given A.P.
Solution 20
The first three digit number which is divisible by 7 is 105 and the last digit which is divisible by 7 is 994.
This is an A.P. in which a = 105, d = 7 and tn = 994.
We know that nth term of A.P is given by
tn = a + (n - 1)d.
⇒ 994 = 105 + (n - 1)7
⇒ 889 = 7n - 7
⇒ 896 = 7n
⇒ n = 128
∴ There are 128 three digit numbers which are divisible by 7.
Solution 21
Let the three parts of 216 in A.P be (a - d), a, (a + d).
⇒a - d + a + a + d = 216
⇒ 3a = 216
⇒ a = 72
Given that the product of the two smaller parts is 5040.
⇒a(a - d ) = 5040
⇒ 72(72 - d) = 5040
⇒ 72 - d = 70
⇒ d = 2
∴ a - d = 72 - 2 = 70, a = 72 and a + d = 72 + 2 = 74
Therefore the three parts of 216 are 70, 72 and 74.
Solution 22
We have 2n2 - 7,
Substitute n = 1, 2, 3, … , we get
2(1)2 - 7, 2(2)2 - 7, 2(3)2 - 7, 2(4)2 - 7, ….
-5, 1, 11, ….
Difference between the first and second term = 1 - (-5) = 6
And Difference between the second and third term = 11 - 1 = 10
Here, the common difference is not same.
Therefore the nth term of an A.P can't be 2n2 - 7.
Solution 23
Here a = 14 , d = 7 and tn = 168
tn = a + (n - 1)d
⇒ 168 = 14 + (n - 1)7
⇒ 154 = 7n - 7
⇒ 154 = 7n - 7
⇒ 161 = 7n
⇒ n = 23
We know that,
Therefore the sum of the A.P., 14, 21, 28, …, 168 is 2093.
Solution 24
Here a = 20 and S7 = 2100
We know that,
To find: t31 =?
tn = a + (n - 1)d
Therefore the 31st term of the given A.P. is 2820.
Solution 25
First we will reverse the given A.P. as we have to find the sum of last 8 terms of the A.P.
58, …., -8, -10, -12.
Here a = 58 , d = -2
Therefore the sum of last 8 terms of the A.P. -12, -10, -8, ……, 58 is 408.
