NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.3
Question 31

Question. 31

Find the sum of first seven numbers which are multiples of 2 as well as of 9.

Answer:

504

Detailed Answer with Explanation:

Step 1: We need numbers that are multiples of both 2 and 9.

The smallest number that is a multiple of both 2 and 9 is their Least Common Multiple (LCM).

LCM of 2 and 9 is:

2 = \(2\)

9 = \(3 \times 3\)

LCM = \(2 \times 3 \times 3 = 18\).

Step 2: So, the first number which is a multiple of both 2 and 9 is 18.

The sequence of such numbers will be: \(18, 36, 54, 72, 90, 108, 126, \ldots\)

Step 3: We only need the first 7 terms:

\(18, 36, 54, 72, 90, 108, 126\).

Step 4: This is an Arithmetic Progression (AP) because the difference between terms is constant.

Common difference (d) = \(36 - 18 = 18\).

Step 5: Formula for the sum of first \(n\) terms of an AP is:

\[ S_n = \dfrac{n}{2} \times (a + l) \]

where,

\(n = 7\) (number of terms),

\(a = 18\) (first term),

\(l = 126\) (last term).

Step 6: Substitute the values:

\[ S_7 = \dfrac{7}{2} \times (18 + 126) \]

\[ S_7 = \dfrac{7}{2} \times 144 \]

\[ S_7 = 7 \times 72 \]

\[ S_7 = 504 \]

Final Answer: The sum of the first seven numbers which are multiples of both 2 and 9 is 504.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 5: Arithematic Progressions – Exercise 5.3 | Detailed Answers