Question

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

Answer 1

Correct Answer: 6993

Step-by-step solution

1. Observe the pattern of set sizes.
   The n-th set contains 2n − 1 elements (odd numbers 1,3,5,…).
2. Find how many terms appear before the n^th set.
   The total number of terms in the first (n−1) sets is the sum of the first (n−1) odd numbers: \[ \sum_{k=1}^{\,n-1} (2k-1) = (n-1)^2. \] So the index (in the sequence of multiples of 3) of the first term of the n^th set is \[ \text{first index} = (n-1)^2 + 1. \]
3. Apply the formula for n = 11.
   For the 11^th set: \[ (n-1)^2 = (11-1)^2 = 10^2 = 100, \] so the first term is the 101^st positive multiple of 3: \[ a_1 = 3 \times 101 = 303. \] The number of terms in the 11^th set is \[ 2n-1 = 21. \] These 21 terms are consecutive multiples of 3: \[ 303,\;306,\;309,\;\dots,\;303 + (21-1)\cdot 3. \] The last term is \[ a_{21} = 303 + 60 = 363. \]
4. Compute the sum of this arithmetic progression.
   The sum of an arithmetic progression with \(\,N\) terms, first term \(a\) and last term \(l\) is \[ S = \frac{N}{2}\,(a + l). \] Here \(N=21,\; a=303,\; l=363\). So \[ S = \frac{21}{2}\,(303 + 363) = \frac{21}{2}\cdot 666 = 21\cdot 333 = 6993. \]

Final answer

The sum of the elements in the 11^th set is 6993.

Bonus — compact general formula

For the n-th set (with size \(2n-1\)), the middle term index is \((n-1)^2 + n\), so the middle term value is \[ 3\big((n-1)^2 + n\big) = 3\big(n^2 - n + 1\big). \] Since an arithmetic block of odd length has the average equal to the middle term, the sum of the n^th set is \[ S_n = (2n-1)\times 3\big(n^2-n+1\big) = 3(2n-1)(n^2-n+1). \] For \(n=11\) this gives \(S_{11} = 3\cdot 21 \cdot 111 = 6993\), confirming the result.