Let the common difference = d

S = $$\sum\limits_{i = 1}^{30} {{a_i}} $$

= $$a$$₁ + $$a$$₂ + . . . . . + $$a$$₃₀

$$ \therefore $$  S = $${{30} \over 2}\left[ {{a_1} + {a_{30}}} \right]$$

= 15 [$$a$$₁ + $$a$$₁ + 29d]

= 15 (2$$a$$₁ + 29d)

T = $$\sum\limits_{i = 1}^{15} {{a_{\left( {2i - 1} \right)}}} $$

= $$a$$₁ + $$a$$₃ + . . . . . . + $$a$$₂₉

= $${{15} \over 2}\left[ {a{}_1 + {a_{29}}} \right]$$

= $${{15} \over 2}\left[ {a{}_1 + {a_1} + 28d} \right]$$

= $${{15} \over 2}\left[ {2a{}_1 + 28d} \right]$$

= 15 ($$a$$₁ + 14d)

Given,

S $$-$$ 2T = 75

$$ \Rightarrow $$  15(2$$a$$₁ + 29d) $$-$$ 2 $$ \times $$ 15 ($$a$$₁ + 14d) = 75

$$ \Rightarrow $$  30$$a$$₁ + 15 $$ \times $$ 29d $$-$$ 30 $$a$$₁ $$-$$ 420d = 75

$$ \Rightarrow $$  435d $$-$$ 420d = 75

$$ \Rightarrow $$  15d = 75

$$ \Rightarrow $$  d = 5

Given that,

$$a$$₅ = 27

$$ \Rightarrow $$  $$a$$₁ + 4d = 27

$$ \Rightarrow $$  $$a$$₁ + 20 = 27

$$ \Rightarrow $$  $$a$$₁ = 7

$$ \therefore $$  $$a$$₁₀ = $$a$$₁ + 9d

= 7 + 45

= 52