Question:

**Step 1:** Express $S$ and $T$ in terms of the first term $a_1$ and common difference $d$. The sum of an arithmetic progression with $n$ terms is given by $S_n = \frac{n}{2}(2a_1 + (n-1)d)$. For $S = a_1 + a_2 + \cdots + a_{30}$: $$S = \frac{30}{2}(2a_1 + (30-1)d) = 15(2a_1 + 29d) = 30a_1 + 435d$$ For $T = \displaystyle\sum_{i=1}^{15} a_{(2i-1)} = a_1 + a_3 + \cdots + a_{29}$: This is an arithmetic progression with 15 terms. The first term is $a_1$. The common difference between consecutive terms in this sum is $a_3 - a_1 = (a_1+2d) - a_1 = 2d$. $$T = \frac{15}{2}(2a_1 + (15-1)(2d)) = \frac{15}{2}(2a_1 + 14 \cdot 2d) = 15(a_1 + 14d) = 15a_1 + 210d$$ **Step 2:** Use the given condition $S - 2T = 75$. Substitute the expressions for $S$ and $T$: $$(30a_1 + 435d) - 2(15a_1 + 210d) = 75$$ $$30a_1 + 435d - 30a_1 - 420d = 75$$ $$15d = 75$$ $$d = 5$$ **Step 3:** Use the given condition $a_5 = 27$. The $k$-th term of an arithmetic progression is $a_k = a_1 + (k-1)d$. For $a_5 = 27$: $$a_1 + (5-1)d = 27$$ $$a_1 + 4d = 27$$ **Step 4:** Solve for $a_1$. Substitute the value of $d=5$ into the equation from Step 3: $$a_1 + 4(5) = 27$$ $$a_1 + 20 = 27$$ $$a_1 = 7$$ **Step 5:** Calculate $a_{10}$. Using the formula $a_k = a_1 + (k-1)d$ for $a_{10}$: $$a_{10} = a_1 + (10-1)d$$ $$a_{10} = a_1 + 9d$$ Substitute $a_1=7$ and $d=5$: $$a_{10} = 7 + 9(5)$$ $$a_{10} = 7 + 45$$ $$a_{10} = 52$$