Question:

The sum of first 5 multiples of 3 is :

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This problem can be solved by direct summation for a small number of terms. For a larger number of terms, it's more efficient to recognize that multiples of a number form an arithmetic progression and use the sum formula for an AP.
Updated On: Jun 5, 2025
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The Correct Option is A

Solution and Explanation

Step 1: Identify the first 5 multiples of 3.
The multiples of 3 are numbers that can be obtained by multiplying 3 by an integer.
The first 5 multiples of 3 are:
$3 \times 1 = 3$
$3 \times 2 = 6$
$3 \times 3 = 9$
$3 \times 4 = 12$
$3 \times 5 = 15$
So, the multiples are 3, 6, 9, 12, 15.
Step 2: Calculate the sum of these multiples.
Sum = $3 + 6 + 9 + 12 + 15$ Method 1: Direct Summation
Sum = $3 + 6 + 9 + 12 + 15 = 45$ Method 2: Using Arithmetic Progression (AP) formula
The multiples of 3 form an arithmetic progression with:
First term ($a$) = 3
Common difference ($d$) = 3
Number of terms ($n$) = 5 The sum of an arithmetic progression is given by the formula:
$S_n = \frac{n}{2}(2a + (n-1)d)$
or
$S_n = \frac{n}{2}(a + l)$, where $l$ is the last term.
Using the first formula:
$S_5 = \frac{5}{2}(2 \times 3 + (5-1) \times 3)$
$S_5 = \frac{5}{2}(6 + 4 \times 3)$
$S_5 = \frac{5}{2}(6 + 12)$
$S_5 = \frac{5}{2}(18)$
$S_5 = 5 \times 9$
$S_5 = 45$
Using the second formula (where $l=15$):
$S_5 = \frac{5}{2}(3 + 15)$
$S_5 = \frac{5}{2}(18)$
$S_5 = 5 \times 9$
$S_5 = 45$
Step 3: Conclude the result.
The sum of the first 5 multiples of 3 is 45. Step 4: Compare with the given options.
The calculated sum is 45, which matches option (1). (1) 45
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