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Q13:
Find the sum of the first 15 multiples of 8.

Solution :

Given:

The sequence consists of the first 15 multiples of 8. This forms an Arithmetic Progression (AP) where:

  • The first term ($a$) = $8 \times 1 = 8$
  • The second term = $8 \times 2 = 16$
  • The common difference ($d$) = $16 - 8 = 8$
  • The number of terms ($n$) = $15$

To Find:

The sum of the first 15 multiples of 8 ($S_{15}$).

Step 1: Identifying the Arithmetic Progression

The sequence of the first 15 multiples of 8 is: $8, 16, 24, 32, \dots$

Here, the first term $a = 8$.

The common difference $d = a_2 - a_1 = 16 - 8 = 8$.

The number of terms $n = 15$.

Step 2: Selecting the Formula

To find the sum of the first $n$ terms of an Arithmetic Progression, we use the formula:

$S_n = \frac{n}{2} [2a + (n - 1)d]$

[Where $S_n$ is the sum of $n$ terms, $n$ is the number of terms, $a$ is the first term, and $d$ is the common difference.]

Step 3: Substituting the Values

Substitute $n = 15$, $a = 8$, and $d = 8$ into the formula:

$S_{15} = \frac{15}{2} [2(8) + (15 - 1)8]$

Step 4: Performing Arithmetic Calculations

First, simplify the expression inside the parentheses:

$S_{15} = \frac{15}{2} [16 + (14)8]$

[Since $15 - 1 = 14$]

Next, calculate the product inside the bracket:

$S_{15} = \frac{15}{2} [16 + 112]$

[Since $14 \times 8 = 112$]

Add the terms inside the bracket:

$S_{15} = \frac{15}{2} [128]$

[Since $16 + 112 = 128$]

Step 5: Final Simplification

$S_{15} = 15 \times \frac{128}{2}$

$S_{15} = 15 \times 64$

[Since $128 \div 2 = 64$]

$S_{15} = 960$

[Since $15 \times 60 = 900$ and $15 \times 4 = 60$, $900 + 60 = 960$]

Final Answer: The sum of the first 15 multiples of 8 is 960.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


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