Let the A.P. have first term $a$ and common difference $d$.

Then $a_n=a+(n-1)d$

We are given:

1. $a_5=27$
2. $S=\sum_{i=1}^{30} a_i$
3. $T=\sum_{i=1}^{15} a_{2i-1}$
4. $S-2T=75$

We need to find $a_{10}$.

1. Use $a_5=27$

Since $a_5=a+4d=27$ we get $a+4d=27 \quad ...(1)$

2. Compute $S$

Sum of first $30$ terms of an A.P. is $S=\frac{30}{2}[2a+(30-1)d]=15(2a+29d)$ So, $S=30a+435d$

3. Compute $T$

The odd-positioned terms are: $a_1,a_3,a_5,\dots,a_{29}$ This is itself an A.P. with:

• first term $a_1=a$
• common difference $2d$
• number of terms $15$

Hence, $T=\frac{15}{2}[2a+(15-1)(2d)]$ $T=\frac{15}{2}(2a+28d)$ $T=15(a+14d)$ $T=15a+210d$

Therefore, $2T=30a+420d$

4. Use the condition $S-2T=75$

Substitute: $S-2T=(30a+435d)-(30a+420d)=15d$ Given that $S-2T=75$ so, $15d=75$ $d=5$

5. Find $a$

From equation (1): $a+4d=27$ $a+20=27$ $a=7$

6. Find $a_{10}$

$a_{10}=a+9d=7+9\cdot 5=7+45=52$

7. Check options

• A: $47$
• B: $42$
• C: $52$ ✅
• D: $57$

Thus, the correct answer is: $\boxed{52}$