NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.4
Question 5

Question. 5

Find the sum of integers between 100 and 200 that are

(i) divisible by 9,   (ii) not divisible by 9.

Answer:

(i) 1683   (ii) 13167

Detailed Answer with Explanation:

Step (i): Integers divisible by 9

1. First, write down the numbers between 100 and 200 which are divisible by 9.

The smallest number greater than 100 divisible by 9 is 108, and the largest below 200 is 198.

So the numbers are: 108, 117, 126, …, 198.

2. This is an arithmetic progression (AP) with first term \(a = 108\), last term \(l = 198\), and common difference \(d = 9\).

3. To find how many terms there are, use the AP formula: \(n = \dfrac{l - a}{d} + 1\).

\(n = \dfrac{198 - 108}{9} + 1 = \dfrac{90}{9} + 1 = 10 + 1 = 11\).

4. The sum of an AP is given by: \(S = \dfrac{n}{2}(a + l)\).

\(S = \dfrac{11}{2}(108 + 198) = \dfrac{11}{2}(306) = 11 \times 153 = 1683\).

So, the sum of numbers divisible by 9 is 1683.


Step (ii): Integers not divisible by 9

1. Integers strictly between 100 and 200 are: 101, 102, …, 199.

This is another AP with first term \(a = 101\), last term \(l = 199\), and number of terms \(n = 99\).

2. The sum of these 99 numbers is:

\(S = \dfrac{n}{2}(a + l) = \dfrac{99}{2}(101 + 199) = \dfrac{99}{2}(300) = 99 \times 150 = 14850\).

3. Out of this total sum, the part divisible by 9 (already found in step (i)) is 1683.

4. Therefore, the sum of integers not divisible by 9 is:

\(14850 - 1683 = 13167\).

So, the sum of numbers not divisible by 9 is 13167.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 5: Arithematic Progressions – Exercise 5.4 | Detailed Answers