Here you get the CBSE Class 10 Mathematics chapter 5, Arithmetic Progressions: NCERT Exemplar Problems and Solutions (Part-IIIC). This part of the chapter includes solutions of Question Number 20 to 29 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.
NCERT Exemplar Solution for CBSE Class 10 Mathematics: Arithmetic Progressions (Part-IIIA)
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.
NCERT Exemplar Solution for CBSE Class 10 Mathematics: Arithmetic Progressions (Part-IIIB)
Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Arithmetic Progressions:
Exercise 5.3
Short Answer Type Questions (Q. No. 20-29):
Question. 20 The first term of an AP is −5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
Solution.
Let the first term, common difference and the number of terms of an AP are a, d and n respectively.
Given that, first term of the AP, a = −5
Last term of AP, l = 45
And, sum of all terms of AP, S_{n} = 120
We know that, if last term of an AP is known, then sum of n terms of an AP is,
Hence, number of terms and the common difference of an AP are 6 and 10 respectively.
Question. 21 Find the sum:
Solution.
Question. 22 Which term of the AP −2, −7, −12, ... Will be −77? Find the sum of this AP upto the term −77.
Solution.
Let −77 be the nth term of AP −2, −7, −12,….
Here, first term, a = − 2
Common difference, d = −7 – (−2) = −7 + 2 = −5
Hence, the sum of the terms upto – 77 of given AP is − 632.
Question. 23 If a_{n} = 3 – 4n, then show that a_{1}, a_{2}, a_{3}, .... form an AP. Also, find S_{20}.
Solution.
Question. 24 In an AP, if S_{n} = n (4n + 1), then find the AP.
Solution.
Given, S_{n} = n (4n + 1) ….(i)
We know that, the nth term if an AP is given as:
a_{n} = S_{n} − S_{n−1}
⟹ a_{n} = n (4n + 1) – (n – 1) [4(n – 1) + 1] [Using (i)]
⟹ a_{n} = 4n^{2} + n – (n – 1) (4n – 3)
⟹ a_{n} = 4n^{2} + n – 4n^{2} + 4n + 3n – 3
⟹ a_{n} = 8n − 3
Put n = 1, a_{1} = 8(1) – 3 = 5
Put n = 2, a_{2} = 8(2) – 3 = 16 – 3 = 13
Put n = 3, a_{3} = 8(3) – 3 = 24 – 3 = 21
Hence, the required AP is 5, 13, 21, .....
Question. 25 In an AP, if S_{n} = 3n^{2} + 5n and a_{k} = 164, then find the value of k.
Solution.
Given, S_{n} = 3n^{2} + 5n …(i)
And a_{k} = 164
We know that, the nth term if an AP is given as:
a_{n} = S_{n} − S_{n−1}
a_{n} = 3n^{2} + 5n – 3(n −1)^{2} – 5(n – 1) [Using (i)]
a_{n} = 3n^{2} + 5n – 3(n^{2} + 1 – 2n) – 5n +5
a_{n} = 3n^{2} + 5n – 3n^{2} – 3 +6n – 5n +5
a_{n} = 6n + 2
Putting n = k in above equation and using equation (ii) , we get:
a_{k} = 6k + 2 = 164
⟹ 6k = 162
⟹ k = 27
Question. 26 If S_{n} denotes the sum of first n terms of an AP, Then Prove that S_{12} = 3 (S_{8} – S_{4}).
Solution.
Question. 27 Find the sum of first 17 terms of an AP whose 4^{th} and 9^{th} terms are – 15 and – 30, respectively.
Solution.
Let the first term, common difference and the number of terms of AP be a, d and n, respectively.
Using formula for nth term of an AP, we have:
a_{n} = a + (n − 1) d
⟹ 4^{th} term of AP, a_{4 }= a + 3d
⟹ a + 3d = −15 [Given]….(i)
Also, 9^{th} term of AP, a_{9} = a+ 8d
⟹ a+ 8d = − 30 [Given]….(ii)
Subtracting equation (ii) from equation (iii), We get
Hence, the required sum of first 17 terms of an AP is −510.
Question. 28 If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, then find the sum of first 10 terms.
Solution.
Let the first term and common difference of AP be a and d respectively.
Given that, sum of first 6 terms of AP, S_{6} = 36
And, sum of first 16 terms of AP, S_{16} = 256
As sum of first n terms of an AP is given as:
Question. 29 Find the sum of all the 11 terms of an AP whose middle most term is 30.
Solution.
Since the total number of terms, n = 11, which is odd.
Hence, the sum of all the 11 terms of AP is 330.
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