Key Concepts and Formulas

• Sum of an Arithmetic Progression (A.P.): The sum of the first $n$ terms of an A.P. with first term $a$ and common difference $d$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
• Summation Properties:
  • $\sum_{i=1}^{N} (a_i - b_i) = \sum_{i=1}^{N} a_i - \sum_{i=1}^{N} b_i$
  • $\sum_{i=1}^{N} c \cdot a_i = c \cdot \sum_{i=1}^{N} a_i$
• Standard Summation Formulas:
  • Sum of the first $N$ natural numbers: $\sum_{i=1}^{N} i = \frac{N(N+1)}{2}$
  • Sum of a constant: $\sum_{i=1}^{N} C = N \cdot C$

Step-by-Step Solution

Step 1: Define the parameters for the $i$-th A.P. We are given 10 A.P.s. Let the $i$-th A.P. (where $i \in \{1, 2, \ldots, 10\}$) have first term $a_i$ and common difference $d_i$. We need to find the sum of the first 12 terms of each, denoted by $s_i$.

• The first terms are $1, 2, 3, \ldots, 10$. Thus, for the $i$-th A.P., $a_i = i$.
• The common differences are $1, 3, 5, \ldots, 19$. This is an arithmetic progression of odd numbers. The $i$-th term of this sequence is given by $2i-1$. Thus, for the $i$-th A.P., $d_i = 2i-1$.
• The number of terms to sum for each A.P. is $n=12$.

Step 2: Derive the formula for $s_i$, the sum of the first 12 terms of the $i$-th A.P. We use the formula for the sum of an A.P., $S_n = \frac{n}{2}[2a + (n-1)d]$, with $n=12$, $a=a_i=i$, and $d=d_i=2i-1$. $s_i = \frac{12}{2}[2(a_i) + (12-1)(d_i)]$ Substitute the expressions for $a_i$ and $d_i$: $s_i = 6[2(i) + 11(2i-1)]$ Simplify the expression inside the brackets: $s_i = 6[2i + 22i - 11]$ $s_i = 6[24i - 11]$ Distribute the 6: $s_i = 144i - 66$ This formula gives the sum of the first 12 terms for the $i$-th A.P.

Step 3: Calculate the total sum $\sum_{i=1}^{10} s_i$. We need to find the sum of $s_i$ for $i$ from 1 to 10. $\sum_{i=1}^{10} s_i = \sum_{i=1}^{10} (144i - 66)$ Using the properties of summation, we can split this into two separate sums: $\sum_{i=1}^{10} s_i = \sum_{i=1}^{10} (144i) - \sum_{i=1}^{10} (66)$ Factor out the constants: $\sum_{i=1}^{10} s_i = 144 \sum_{i=1}^{10} i - 66 \sum_{i=1}^{10} 1$

Step 4: Apply standard summation formulas and compute the final result. We use the formula for the sum of the first $N$ natural numbers, $\sum_{i=1}^{N} i = \frac{N(N+1)}{2}$, with $N=10$: $\sum_{i=1}^{10} i = \frac{10(10+1)}{2} = \frac{10 \times 11}{2} = 55$ And the formula for the sum of a constant, $\sum_{i=1}^{N} 1 = N$: $\sum_{i=1}^{10} 1 = 10$ Now substitute these values back into the expression for the total sum: $\sum_{i=1}^{10} s_i = 144(55) - 66(10)$ Perform the multiplications: $144 \times 55 = 7920$ $66 \times 10 = 660$ Finally, subtract: $\sum_{i=1}^{10} s_i = 7920 - 660 = 7260$

Common Mistakes & Tips

• Misinterpreting Common Differences: Ensure the common difference for the $i$-th AP is correctly identified as $2i-1$, not simply $i$.
• Algebraic Errors: Be careful with distributing terms and simplifying the expression for $s_i$. A small error can lead to a significantly different final answer.
• Summation Formula Accuracy: Double-check the standard summation formulas for the sum of the first $N$ integers and the sum of a constant.

Summary

The problem requires us to calculate the sum of the first 12 terms for 10 different arithmetic progressions. We first establish a general formula for the sum of the $i$-th A.P., $s_i = 144i - 66$. Then, we sum these individual sums from $i=1$ to $10$ using properties of summation and standard summation formulas. This process leads to the total sum of 7260.

The final answer is $\boxed{7260}$.