Here you get the CBSE Class 10 Mathematics chapter 5, Arithmetic Progressions: NCERT Exemplar Problems and Solutions (Part-IIIA). This part of the chapter includes solutions of Question Number 1 to 8 from Exercise 5.3 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Arithmetic Progressions. This exercise comprises of only the Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.
CBSE Class 10 Mathematics Syllabus 2017-2018
NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.
Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Arithmetic Progressions:
Exercise 5.3
Short Answer Type Questions
Question. 1 Match the AP's given in column A with suitable common differences given in column B.
Solution.
(A_{1}) 2, -2, -6, -10,….
Here, common difference, d = - 2 - 2 = - 4
Hence, A_{1} matches to B_{4}.
(A_{2}) a = −18, n = 10, a_{n} = 0
Using, a_{n} = a + (n – 1)d, we get:
0 = −18 + (10 − 1)d
⟹ 18 = 9d
⟹ d = 2
Hence, A_{2} matches to B_{5}.
(A_{3}) a = 0, a_{10} = 6
Hence, A_{3 }matches to B_{1}.
(A_{4}) a_{2} = 13, a_{4} = 3
Using, a_{n} = a + (n – 1)d, we get:
a_{2} = a + (2 − 1)d
⟹ a + d = 13 …(i)
Again, a_{4} = a + (4 − 1)d
⟹ a + 3d = 3 ….(ii)
Subtracting equation (i) from equation (ii), we get:
2d = −10
⟹ d = − 5
Hence, A_{4} matches to B_{2}.
Question. 2 Verify that each of the following is an AP and Then Write its next three terms.
Solution.
A series of numbers will form an AP, if d_{1} = d_{2 }= d_{3}…
(iv) a + b, (a + 1) + b, (a + 1) + (b + 1), ….
Here, a_{1} = a + b, a_{2} = (a + 1) + b, a_{3} = (a + 1) + (b + 1)
Now, d_{1} = a_{2} – a_{1} = [(a + 1) + b] – (a + b) = a + 1 + b – a – b = 1
d_{2} = a_{3} – a_{2} = [(a + 1) + (b + 1)] – [(a + 1) + b]
= a + 1 + b + 1 – a – 1 – b = 1
As d_{1} = d_{2}, therefore given list of numbers forms an AP.
Using, a_{n} = a + (n – 1)d, next three term are calculated as:
a_{4} = a_{1} + 3d = a + b + 3(1) = (a+2) + (b+1)
a_{5} = a_{1} + 4d = a + b + 4(1) = (a+2) + (b+2)
a_{6} = a_{1} + 5d = a + b + 5(1) = (a+3) + (b+2)
(v) a, 2a + 1, 3a + 2, 4a + 3, ….
Here, a_{1} = a, a_{2} = 2a + 1, a_{3} = 3a + 2 and a_{4} = 4a + 3
Now, d_{1} = a_{2} – a_{1} = 2a + 1 – a = a + 1
d_{2} = a_{3} – a_{2} = 3a + 2 – 2a – 1 = a + 1
d_{3} = a_{4} – a_{3} = 4a + 3 – 3a – 2 = a + 1
As d_{1} = d_{2 }= d_{3}, therefore given list of numbers forms an AP.
Using, a_{n} = a + (n – 1)d, next three term are calculated as:
a_{5} = a_{1} + 4d = a + 4 (a+1) = 5a + 4
a_{6} = a_{1} + 5d = a + 5 (a+1) = 6a + 5
a_{7} = a_{1} + 6d = a + 6 (a+1) = 7a + 6
Question. 3 Write the first three terms of the AP’s, When a and d are as given below
(ii) Given, first term, a = − 5
Common difference, d = − 3
Using, a_{n} = a + (n – 1)d, first three term of AP are calculated as:
Second term of AP, a_{2} = a + d = −5 – 3 = −8
And third term of AP, a_{3} = a + 2d = −5 + 2 (−3)
= −5 – 6 = −11
Hence, required three terms are −5, −8, and −11.
Question. 4 Find a, b and c such that the following numbers are in AP:
a, 7, b, 23 and c.
Solution.
Here, d_{1} = 7 – a ….(i)
d_{2} = b – 7 ….(ii)
d_{3} = 23 – b ….(iii)
d_{4} = c – 23 ….(iv)
Since a, 7, b, 23 and c are in AP.
Therefore, d_{1} = d_{2} = d_{3} = d_{4}
Taking (ii) and (iii), we get
b − 7 = 23 – b
⟹ 2b = 30
⟹ b = 15
Taking (i) and (ii), we get
7 – a = b – 7
⟹ 7 – a = 15 – 7 [Using b = 15]
⟹ 7 – a = 8
⟹ a = − 1
Taking (iii) and (iv), we get
23 – b = c – 23
⟹ 23 – 15 = c – 23 [Using b = 15]
⟹ 8 = c – 23
⟹ 8 + 23 = c
⟹ c = 31
Hence, a = − 1, b = 15, c = 31.
Question. 5 Determine the AP whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20.
Solution.
Let the first term of AP be a and common difference d.
Using, a_{n} = a + (n – 1)d, we have:
a_{5} = a + (5 – 1) d = 19 [Given, a_{5} = 19]
⟹ a + 4d = 19 ….(i)
Also, a_{8} = a + (8 – 1) d = a + 7d
And, a_{13} = a + (13 – 1) d = a + 12d
According to question:
a_{13 }− a_{8} = 20
⟹ (a + 12d) – (a + 7d) = 20
⟹ 5d = 20
⟹ d = 4
Putting d = 4 in equation (i), we get
a + 4(4) = 19
⟹ a + 16 = 19
⟹ a = 19 – 16 = 3
So, required AP is given as: a, a + d, a + 2d, a + 3d, ... = 3, 3 + 4, 3 + 2(4), 3 + 3(4), …= 3, 7, 11, 15, .....
Question. 6 The 26^{th}, 11^{th} and the last terms of an AP are, 0,3 and −1/5, respectively. Find the common difference and the number of terms.
Solution.
Let the first term, common difference and number of terms of AP be a, d and n, respectively.
Hence, the common difference and number of terms are – 1/ 5 and 27, respectively.
Question. 7 The sum of the 5th and the 7th terms of an AP is 52 and the 10^{th} term is 46. Find the AP.
Solution.
Let the first term and common difference of AP be a and d, respectively.
According to the question,
a_{5} + a_{7 }= 52 and a_{10} = 46
⟹ (a + 4d) + (a + 6d) = 52 and a + 9d = 46 [Using, a_{n} = a + (n – 1)d]
⟹ 2a + 10d = 52
Or a + 5d = 26 ….(i)
And a + 9d = 46 ….(ii)
Subtracting equation (i) from equation (ii), we get:
4d = 20
⟹ d = 5
Putting d = 5, in equation (i) we get:
a + 5 (5) = 26
⟹ a = 1
So, Required AP is given as: a, a + d, a + 2d, a + 3d, ... = 1, 1 + 5, 1 + 2 (5), 1 + 3(5), .... = 1, 6, 11, 16, .....
Question. 8 Find the 20th term of the AP whose 7th term is 24 less than the 11th term, first term being 12.
Solution.
Let the first term, common difference and number of terms of an AP are a, d and n, respectively.
Given that, first term, a = 12.
According to question,
a_{7} = a_{11} – 24
⟹ a + 6d = (a + 10d) – 24 [Using, a_{n} = a + (n – 1)d]
⟹ 4d = 24
⟹ d = 6
Hence, 20th term of the AP, a_{20} = a + 19d = 12 + 19(6) = 126
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