Key Concepts and Formulas

• Mean (Average) of a Set of Numbers: The mean of $N$ numbers ($x_1, x_2, \ldots, x_N$) is the sum of all numbers divided by the total count of numbers. $\bar{x} = \frac{\sum_{i=1}^{N} x_i}{N}$
• Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant. The $n$-th term ($a_n$) of an AP with first term $a_1$ and common difference $d$ is given by $a_n = a_1 + (n-1)d$.
• Summation Formulas: For the first $k$ natural numbers:
  • Sum of $n$: $\sum_{n=1}^{k} n = \frac{k(k+1)}{2}$
  • Sum of $n^2$: $\sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}$
  • Sum of a constant $c$: $\sum_{n=1}^{k} c = ck$

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Step-by-Step Solution

Step 1: Identify the pattern and determine the general term ($T_n$) of the sequence.

The given sequence of numbers is: $7 \times 8$, $10 \times 10$, $13 \times 12$, $16 \times 14$, ... We need to find a general formula for the $n$-th term, $T_n$. To do this, we analyze the patterns of the first and second factors separately.

1. Sequence of First Factors: 7, 10, 13, 16, ...

   • We check the differences between consecutive terms: $10-7=3$, $13-10=3$, $16-13=3$.
   • Since the common difference is constant, this is an Arithmetic Progression (AP).
   • The first term ($a_1$) is 7.
   • The common difference ($d_1$) is 3.
   • Using the AP formula $a_n = a_1 + (n-1)d$, the $n$-th first factor is $F_n = 7 + (n-1)3 = 7 + 3n - 3 = 3n + 4$.

2. Sequence of Second Factors: 8, 10, 12, 14, ...

   • We check the differences between consecutive terms: $10-8=2$, $12-10=2$, $14-12=2$.
   • This is also an Arithmetic Progression (AP).
   • The first term ($b_1$) is 8.
   • The common difference ($d_2$) is 2.
   • Using the AP formula $b_n = b_1 + (n-1)d$, the $n$-th second factor is $S_n = 8 + (n-1)2 = 8 + 2n - 2 = 2n + 6$.

Each term in the original sequence is the product of its corresponding first and second factors. Therefore, the general $n$-th term ($T_n$) is: $T_n = F_n \times S_n = (3n + 4)(2n + 6)$ This formula will generate any term in the sequence. For example, for $n=1$, $T_1 = (3(1)+4)(2(1)+6) = (7)(8) = 56$, which matches the first given term.

Step 2: Simplify the general term ($T_n$) into a polynomial form.

To easily sum the terms, we expand the expression for $T_n$: $T_n = (3n + 4)(2n + 6)$ We can factor out a 2 from the second binomial to simplify: $T_n = (3n + 4) \cdot 2(n + 3)$ $T_n = 2(3n + 4)(n + 3)$ Now, expand the product of the two binomials: $T_n = 2(3n \cdot n + 3n \cdot 3 + 4 \cdot n + 4 \cdot 3)$ $T_n = 2(3n^2 + 9n + 4n + 12)$ Combine the like terms ($9n + 4n = 13n$): $T_n = 2(3n^2 + 13n + 12)$ Finally, distribute the 2: $T_n = 6n^2 + 26n + 24$ This polynomial form is suitable for applying summation formulas.

Step 3: Calculate the sum of the first 10 terms ($S_{10}$).

We need to find the sum of the first 10 terms, $S_{10} = \sum_{n=1}^{10} T_n$. Substitute the simplified expression for $T_n$: $S_{10} = \sum_{n=1}^{10} (6n^2 + 26n + 24)$ Using the linearity property of summation, we can split this into three separate sums: $S_{10} = 6 \sum_{n=1}^{10} n^2 + 26 \sum_{n=1}^{10} n + 24 \sum_{n=1}^{10} 1$ Now, apply the standard summation formulas for $k=10$:

• $\sum_{n=1}^{10} n^2 = \frac{10(10+1)(2 \cdot 10+1)}{6} = \frac{10 \cdot 11 \cdot 21}{6} = \frac{2310}{6} = 385$
• $\sum_{n=1}^{10} n = \frac{10(10+1)}{2} = \frac{10 \cdot 11}{2} = \frac{110}{2} = 55$
• $\sum_{n=1}^{10} 1 = 10 \cdot 1 = 10$

Substitute these values back into the expression for $S_{10}$: $S_{10} = 6(385) + 26(55) + 24(10)$ Perform the multiplications: $S_{10} = 2310 + 1430 + 240$ Add these values to find the total sum: $S_{10} = 3740 + 240$ $S_{10} = 3980$ The sum of the first 10 numbers in the sequence is 3980.

Step 4: Calculate the mean of the 10 numbers.

We have the sum of the 10 numbers ($S_{10} = 3980$) and the total count of numbers ($N=10$). Using the mean formula: $\text{Mean} = \frac{S_{10}}{N}$ $\text{Mean} = \frac{3980}{10}$ $\text{Mean} = 398$

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Common Mistakes & Tips

• Pattern Recognition Accuracy: Ensure you correctly identify the type of sequence (AP, GP, etc.) for both factors and accurately determine their first term and common difference/ratio. Errors here will propagate throughout the entire solution.
• Algebraic Precision: Be careful when expanding and simplifying the general term $T_n$. Mistakes in multiplication or combining like terms are common. Factoring out common constants early can sometimes make the algebra less error-prone.
• Summation Formula Application: Memorize the standard summation formulas for $\sum n$ and $\sum n^2$. Double-check that you substitute the correct upper limit ($k=10$ in this case) into the formulas.
• Arithmetic Errors: This problem involves several arithmetic calculations. Break them down into smaller, manageable steps to minimize the chance of calculation mistakes. A reasonableness check at the end (e.g., does a mean of 398 make sense for terms ranging from 56 to 884?) can help catch significant errors.

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Summary

This problem required finding the mean of a sequence of 10 numbers, where each number is a product of two factors that themselves form arithmetic progressions. The solution involved first identifying the general $n$-th term ($T_n$) by analyzing the patterns of its factors. This $T_n$ was then expanded into a polynomial form ($6n^2 + 26n + 24$). Next, standard summation formulas for $\sum n^2$, $\sum n$, and $\sum 1$ were used to calculate the sum of the first 10 terms. Finally, the mean was computed by dividing this total sum by 10.

The final answer is $\boxed{398}$.