 # NCERT Exemplar Solution for CBSE Class 10 Mathematics: Arithmetic Progressions (Part-IVA)

In this article you will get CBSE Class 10 Mathematics chapter 5, Arithmetic Progressions: NCERT Exemplar Problems and Solutions (Part-IVA). Every question has been provided with a detailed solution. All the questions given in this article are very important to prepare for CBSE Class 10 Board Exam 2017-2018.

Created On: Jul 10, 2017 13:01 IST
Modified On: Jul 10, 2017 14:06 IST  Here you get the CBSE Class 10 Mathematics chapter 5, Arithmetic Progressions: NCERT Exemplar Problems and Solutions (Part-IVA). This part of the chapter includes solutions of Question Number 1 to 5 from Exercise 5.4 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Arithmetic Progressions. This exercise comprises of only the Long 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 5.4

Long Answer Type Questions (Q. No. 1-5):

Question. 1 The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the some of the first ten terms of this AP is 235, find the sum of its first twenty terms.

Solution.

Let the first term, common difference and the number of terms of given AP be a, d and n, respectively.   Question. 2 Find the

(i) Sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.

(ii) Sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.

(iii) Sum of those integers from 1 to 500 which are multiples of 2 or 5.

Solution.

(i) We first find the LCM of 2 and 5 which is 10.

Now all those integers which are multiples of 10 are also the multiples of 2 and 5.

Therefore, multiples of 2 as well as of 5 between 1 and 500 are 10, 20, 30, ......, 490 ….(i)

Series (i) forms an AP with first term, a = 10 and common difference, d = 20 – 10 = 10

Let total number of terms of this AP be n.

Therefore, nth term of AP, an = Last term, l = 490 (ii) All those integers which are multiples of 10 (LCM of 2 and 5) are also the multiples of 2 and 5.

Therefore multiples of 2 as well as of 5 from 1 to 500 are 10, 20, 30, ......., 500  ….(i)

Here, (i) forms an AP with,

First term, a = 10

Common difference, d = 10

Let total number of terms of this AP be n.

Therefore, nth term of AP, an = Last term, l = 490 (iii) Since, multiples of 2 or 5 from 1 to 500 = Multiple of 2 from 1 to 500 + Multiples of 5 from 1 to 500 – Multiple of 10 from 1 to 500

= (2, 4, 6, ....., 500) + (5, 10, 15, ....., 500) – (10, 20, ....., 500)          .....(i)

Let AP1 = 2, 4, 6, ....., 500

AP2 = 5, 10, 15, ....., 500

And AP3 = 10, 20, ....., 500

Let number of terms in AP1, AP2 and AP3 be n1, n2 and n3 respectively.

Now formula for last term of an AP is given as:

l = an + (n – 1)d                      ….(ii)

For AP1,

500 = 2 + (n1 – 1)2      ….[Using (ii)]

⟹       250 = 1+ (n1 – 1)

⟹         n1 = 250

For AP2,

500 = 5 + (n2 – 1)5      ….[Using (ii)]

⟹       100 = 1 + (n2 – 1)

⟹          n2 = 100

For AP3,

500 = 10 + (n3 – 1)10  ….[Using (ii)]

⟹       250 = 1 + (n3 − 1)

⟹       n3 = 50

Thus, from equation (i),

Sum of Multiples of 2 or 5 from 1 to 500

= Sum of (2, 4, 6, ...... 500) + Sum of (5, 10, ..... 500) – Sum of (10, 20, ..... 500) Question. 3 The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the 15th term.

Solution.

Let the first term and common difference of AP be a and d, respectively.
According to the first condition, we have:  Question. 4 An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the AP.

Solution.

Given, total number of terms, n = 37, which is odd.  Question. 5 Find the sum of the integers between 100 and 200 that are

(i) Divisible by 9.

(ii) Not divisible by 9.

Solution.

(i) The integers between 100 and 200 which is divisible by 9 are 108, 117, 126, ... 198 which form an AP with first term, a = 108 and common difference,  d = 117 – 108 = 9

Let n be the number of terms between 100 and 200 which is divisible by 9. Hence, required sum of the integers between 100 and 200 that are divisible by 9 is 1683.

(ii) The sum of the integers between 100 and 200 which is not divisible by 9 = (sum of total numbers between 100 and 200) – (Sum of total number between 100 and 200 which is divisible by 9).     ....(i)

For terms between 100 and 200,

First term, a = 101,  Common difference, d = 102 – 101 = 1  and an = l = 199 Using equation (i), Sum of the integers between 100 and 200 which

are not divisible by 9 = 14850 – 1683       [From solution of part (i)]

=13167

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