Key Concepts and Formulas

• General Term of an A.P.: The $n$-th term of an arithmetic progression with first term $a_1$ and common difference $d$ is given by $a_n = a_1 + (n-1)d$.
• Sum of an A.P.: The sum of the first $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}(a_1 + a_n)$ or $S_n = \frac{n}{2}(2a_1 + (n-1)d)$.

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Step-by-Step Solution

Step 1: Express $S$ in terms of $a_1$ and $d$. The sum $S$ is the sum of the first 30 terms of the A.P. Using the formula $S_n = \frac{n}{2}(2a_1 + (n-1)d)$ with $n=30$: $S = \sum_{i=1}^{30} a_i = \frac{30}{2}(2a_1 + (30-1)d)$ $S = 15(2a_1 + 29d)$

Step 2: Express $T$ in terms of $a_1$ and $d$. The sum $T$ is the sum of the odd-indexed terms from $a_1$ to $a_{29}$. $T = \sum_{i=1}^{15} a_{(2i-1)} = a_1 + a_3 + a_5 + \ldots + a_{29}$ The terms $a_1, a_3, a_5, \ldots, a_{29}$ form an A.P. with the first term $a_1$, a common difference of $2d$, and there are 15 terms. Using the formula $S_n = \frac{n}{2}(a_{first} + a_{last})$ for this sub-series, where $n=15$, $a_{first} = a_1$, and $a_{last} = a_{29}$. We know $a_{29} = a_1 + (29-1)d = a_1 + 28d$. So, $T = \frac{15}{2}(a_1 + a_{29})$ $T = \frac{15}{2}(a_1 + (a_1 + 28d))$ $T = \frac{15}{2}(2a_1 + 28d)$ $T = 15(a_1 + 14d)$

Step 3: Use the given relation $S - 2T = 75$ to find $d$. Substitute the expressions for $S$ and $T$ into the given equation: $15(2a_1 + 29d) - 2 \times [15(a_1 + 14d)] = 75$ $15(2a_1 + 29d) - 30(a_1 + 14d) = 75$ Expand the terms: $30a_1 + 435d - 30a_1 - 420d = 75$ The $30a_1$ terms cancel out: $(435 - 420)d = 75$ $15d = 75$ Divide by 15 to find $d$: $d = \frac{75}{15} = 5$ The common difference $d$ is 5.

Step 4: Use $a_5 = 27$ to find $a_1$. We are given that the 5th term of the A.P. is 27. Using the general term formula $a_n = a_1 + (n-1)d$: $a_5 = a_1 + (5-1)d$ $a_5 = a_1 + 4d$ Substitute the given value $a_5 = 27$ and the calculated value $d=5$: $27 = a_1 + 4(5)$ $27 = a_1 + 20$ Solve for $a_1$: $a_1 = 27 - 20 = 7$ The first term $a_1$ is 7.

Step 5: Calculate $a_{10}$. Now that we have $a_1 = 7$ and $d = 5$, we can find the 10th term using the general term formula: $a_{10} = a_1 + (10-1)d$ $a_{10} = a_1 + 9d$ Substitute the values of $a_1$ and $d$: $a_{10} = 7 + 9(5)$ $a_{10} = 7 + 45$ $a_{10} = 52$

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Common Mistakes & Tips

• Confusing the common difference: Be careful when calculating the sum of odd-indexed terms ($T$). Remember that the common difference of the sub-series ($a_1, a_3, \ldots$) is $2d$, not $d$.
• Algebraic errors: Double-check your expansions and subtractions when simplifying the equation $S - 2T = 75$.
• Identifying the number of terms: Ensure you correctly identify the number of terms in both $S$ (which is 30) and $T$ (which is 15).

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Summary

The problem requires us to find the 10th term of an arithmetic progression given some information about the sum of its terms and a specific term's value. We first expressed the given sums $S$ and $T$ in terms of the first term ($a_1$) and the common difference ($d$) of the A.P. By substituting these expressions into the equation $S - 2T = 75$, we were able to solve for the common difference $d$. Subsequently, using the given value of $a_5$, we determined the first term $a_1$. Finally, with both $a_1$ and $d$ known, we calculated the 10th term, $a_{10}$.

The final answer is $\boxed{52}$.