The series of positive multiples of 3 is divided into sets : {3}, {6, 9, 12}, {15, 18, 21, 24, 27},

Q: Why is the number of elements in the $k^{th}$ set $2k-1$?

Look at the sequence of counts: 1, 3, 5... This is an Arithmetic Progression (AP) with the first term $a = 1$ and common difference $d = 2$. The $k^{th}$ term of an AP is given by $a + (k-1)d$. Substituting our values, we get $1 + (k-1)2 = 1 + 2k - 2 = 2k - 1$. It is a beautiful, consistent pattern!

Q: Why do we use $10^2$ to find the starting position?

The total number of elements in the first $n$ sets is the sum of the first $n$ odd numbers, which is $\sum_{k=1}^{n} (2k-1) = n^2$. For the $11^{th}$ set, we need the sum of all elements in the first 10 sets, which is $10^2 = 100$. This means the $11^{th}$ set begins immediately after the $100^{th}$ element.

Q: Is there a faster way than calculating the last term?

Yes! You can use the sum of an AP formula $S = \frac{n}{2}(2a + (n-1)d)$, where $a$ is the first term (303), $n$ is the number of terms (21), and $d$ is the common difference of the multiples of 3 (which is 3). Both methods lead to the same result, but calculating the last term is often more intuitive.

Question

The series of positive multiples of 3 is divided into sets : $\{3\}, \{6, 9, 12\}, \{15, 18, 21, 24, 27\}, \dots$ Then the sum of the elements in the $11^{th}$ set is equal to ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The sequence consists of multiples of 3: $3, 6, 9, 12, 15, \dots$. These are grouped into sets where the $k^{th}$ set contains $n_k = 2k - 1$ elements.

The number of elements in each set follows the sequence of odd numbers: $1, 3, 5, \dots$. This structure allows us to determine the position of any set within the infinite sequence of multiples.

### The Cumulative Leap

To find the sum of the $11^{th}$ set, we must first identify the starting position. We calculate the total number of elements contained in the first 10 sets.

The sum of the first $n$ odd numbers is given by $n^2$. For $n=10$, the total number of elements preceding the $11^{th}$ set is:

$$10^2 = 100$$

This confirms that the $11^{th}$ set begins immediately after the $100^{th}$ multiple of 3. Therefore, the first term of the $11^{th}$ set is the $101^{st}$ multiple of 3.

### The Heart of the Set

The first term $a$ of the $11^{th}$ set is:

$$a = 3 	imes 101 = 303$$

Using the formula $n_k = 2k - 1$, the number of terms in the $11^{th}$ set is:

$$n_{11} = 2(11) - 1 = 21$$

Since the set starts at the $101^{st}$ multiple and contains 21 terms, it ends at the $(100 + 21)^{st} = 121^{st}$ multiple of 3. The last term $l$ is:

$$l = 3 	imes 121 = 363$$

### The Grand Finale

We now treat the $11^{th}$ set as an Arithmetic Progression with $n = 21$ terms, first term $a = 303$, and last term $l = 363$. The sum $S_{11}$ is calculated using the formula:

$$S_n = \frac{n}{2}(a + l)$$

Substituting our values into the equation:

$$S_{11} = \frac{21}{2}(303 + 363)$$

$$S_{11} = \frac{21}{2}(666)$$

$$S_{11} = 21 	imes 333$$

$$S_{11} = 6993$$

The sum of the $11^{th}$ set is **6993**.

Animated Solution for Mathematics - Sequence and Series: The series of positive multiples of 3 is divided into sets : ${3}, {6, 9, 12}, {15, 18, 21, 24, 27}, \dots$ Then the sum of the elements in the $1 1^{t h}$ set is equal to ________.

Enter Numerical Value:

Visualized Solution

Observe the Set Pattern

General Formula for Number of Elements

Total Elements in First $10$ Sets

Identify the Starting Term of Set $11$

Number of Terms in Set $11$

Identify the Last Term of Set $11$

Apply the Sum of AP Formula

Final Calculation

Key Takeaways

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Cumulative Leap

The Heart of the Set

The Grand Finale

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