Key Concepts and Formulas

1. General Term of an A.P.: The $k$-th term of an Arithmetic Progression (A.P.) with first term $a_1$ and common difference $d$ is given by $a_k = a_1 + (k-1)d$.
2. Sum of an A.P.: The sum of $N$ terms of an A.P. is $S_N = \frac{N}{2}(\text{first term} + \text{last term})$. This formula is particularly useful when the first and last terms are known or can be easily determined.
3. Sub-sequences forming an A.P.: If the indices of terms selected from an A.P. form an A.P., then the selected terms themselves also form an A.P.

Step-by-Step Solution

Step 1: Analyze the given sum and extract information. We are given that $a_1, a_2, \dots, a_n$ are in A.P. The given sum is $a_1 + a_4 + a_7 + \dots + a_{16} = 114$. The indices of the terms in this sum are $1, 4, 7, \dots, 16$. We observe that these indices form an A.P. with the first index being $1$ and the common difference of indices being $4 - 1 = 3$. Let's find the number of terms in this sequence of indices. Using the formula for the $k$-th term of an A.P. for the indices: Last Index = First Index + $(N - 1) \times (\text{Common difference of indices})$ $16 = 1 + (N - 1) \times 3$ $15 = (N - 1) \times 3$ $5 = N - 1$ $N = 6$. So, there are 6 terms in the sum $a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16}$. Since the indices form an A.P., the terms $a_1, a_4, a_7, \dots, a_{16}$ themselves form an A.P. We can use the sum formula for an A.P. for these terms: Sum = $\frac{\text{Number of terms}}{2} \times (\text{First term} + \text{Last term})$ $114 = \frac{6}{2} \times (a_1 + a_{16})$ $114 = 3 \times (a_1 + a_{16})$ Dividing both sides by 3, we get: $a_1 + a_{16} = \frac{114}{3}$ $a_1 + a_{16} = 38$.

Step 2: Analyze the required sum. We need to find the value of $a_1 + a_6 + a_{11} + a_{16}$. The indices of the terms in this sum are $1, 6, 11, 16$. We observe that these indices also form an A.P. with the first index being $1$ and the common difference of indices being $6 - 1 = 5$. Let's find the number of terms in this sequence of indices. Last Index = First Index + $(M - 1) \times (\text{Common difference of indices})$ $16 = 1 + (M - 1) \times 5$ $15 = (M - 1) \times 5$ $3 = M - 1$ $M = 4$. So, there are 4 terms in the sum $a_1 + a_6 + a_{11} + a_{16}$. Since the indices form an A.P., the terms $a_1, a_6, a_{11}, a_{16}$ themselves form an A.P. We can use the sum formula for an A.P. for these terms: Required Sum = $\frac{\text{Number of terms}}{2} \times (\text{First term} + \text{Last term})$ Required Sum = $\frac{4}{2} \times (a_1 + a_{16})$ Required Sum = $2 \times (a_1 + a_{16})$.

Step 3: Substitute the result from Step 1 into the expression from Step 2. From Step 1, we found that $a_1 + a_{16} = 38$. Substituting this value into the expression for the required sum from Step 2: Required Sum = $2 \times (38)$ Required Sum = $76$.

Therefore, $a_1 + a_6 + a_{11} + a_{16} = 76$.

Common Mistakes & Tips

• Identify Sub-APs: The key to solving such problems is to recognize that terms with indices in an arithmetic progression also form an arithmetic progression.
• Utilize the Sum Formula: The formula $S_N = \frac{N}{2}(\text{first term} + \text{last term})$ is very efficient when the first and last terms of the sub-sequence are known or can be easily determined.
• Avoid Calculating Individual Terms: It's often unnecessary to find the values of $a_1$ and the common difference $d$ of the original A.P. Focus on relationships between terms, like $a_1 + a_{16}$.

Summary

The problem involves an arithmetic progression where a sum of specific terms is given, and another sum needs to be calculated. By recognizing that the terms involved in both sums form their own arithmetic progressions due to their indices being in an arithmetic progression, we can effectively use the sum formula for an A.P. The first step involved using the given sum to find the value of $(a_1 + a_{16})$. The second step involved using this value to calculate the required sum. This approach avoids the need to find the individual values of $a_1$ and the common difference $d$.

The final answer is $\boxed{76}$.