Key Concepts and Formulas

• General Term of an A.P.: The $n$-th term of an Arithmetic Progression (A.P.) is given by $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference.
• Sum of the First $n$ Terms of an A.P.: The sum of the first $n$ terms, $S_n$, is given by $S_n = \frac{n}{2}(a_1 + a_n)$ or $S_n = \frac{n}{2}(2a_1 + (n-1)d)$.
• Property of Middle Term: For an odd number of terms in an A.P., the sum is equal to the number of terms multiplied by the middle term. For $n$ terms, the middle term is $a_{(n+1)/2}$. Thus, $S_n = n \times a_{(n+1)/2}$ if $n$ is odd.

Step-by-Step Solution

Step 1: Express the given condition in terms of $a_1$ and $d$. We are given the equation $a_1 + a_7 + a_{16} = 40$. Using the formula for the $n$-th term, $a_n = a_1 + (n-1)d$:

• $a_7 = a_1 + (7-1)d = a_1 + 6d$
• $a_{16} = a_1 + (16-1)d = a_1 + 15d$

Substituting these into the given equation: $a_1 + (a_1 + 6d) + (a_1 + 15d) = 40$

Step 2: Simplify the equation to find a relationship between $a_1$ and $d$. Combine the like terms in the equation from Step 1: $(a_1 + a_1 + a_1) + (6d + 15d) = 40$ $3a_1 + 21d = 40$ To simplify further, we can divide the entire equation by 3: $\frac{3a_1}{3} + \frac{21d}{3} = \frac{40}{3}$ $a_1 + 7d = \frac{40}{3}$ Notice that $a_1 + 7d$ is the 8th term of the A.P., i.e., $a_8$. Thus, we have found that $a_8 = \frac{40}{3}$.

Step 3: Calculate the sum of the first 15 terms ($S_{15}$). We need to find $S_{15}$. Using the formula $S_n = \frac{n}{2}(2a_1 + (n-1)d)$, with $n=15$: $S_{15} = \frac{15}{2}(2a_1 + (15-1)d)$ $S_{15} = \frac{15}{2}(2a_1 + 14d)$

Step 4: Simplify the expression for $S_{15}$ and substitute the found relationship. Factor out a 2 from the terms inside the parenthesis: $S_{15} = \frac{15}{2} \cdot 2(a_1 + 7d)$ $S_{15} = 15(a_1 + 7d)$ From Step 2, we know that $a_1 + 7d = \frac{40}{3}$. Substitute this value into the expression for $S_{15}$: $S_{15} = 15 \times \left(\frac{40}{3}\right)$

Step 5: Compute the final value of $S_{15}$. $S_{15} = \frac{15 \times 40}{3}$ $S_{15} = 5 \times 40$ $S_{15} = 200$

Alternatively, using the property of the middle term: Since $n=15$ is odd, the sum of the first 15 terms is $S_{15} = 15 \times a_{(15+1)/2} = 15 \times a_8$. From Step 2, we found $a_8 = \frac{40}{3}$. Therefore, $S_{15} = 15 \times \frac{40}{3} = 5 \times 40 = 200$.

Common Mistakes & Tips

• Algebraic Errors: Be careful with combining terms and simplifying fractions.
• Formula Recall: Ensure you have the correct formulas for the general term and sum of an A.P.
• Recognizing Patterns: The expression $a_1 + 7d$ directly corresponds to $a_8$, and $S_{15} = 15 \times a_8$ is a powerful shortcut if recognized.

Summary

The problem requires us to find the sum of the first 15 terms of an A.P., given a condition involving specific terms. By expressing the given terms in relation to the first term ($a_1$) and common difference ($d$), we derived a simplified equation $a_1 + 7d = \frac{40}{3}$, which directly gives us the value of the 8th term ($a_8$). Using the formula for the sum of the first $n$ terms, or by recognizing that the sum is $n$ times the middle term for an odd $n$, we substituted the value of $a_8$ to find $S_{15} = 200$.

The final answer is $\boxed{200}$, which corresponds to option (B).