Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11^th term is :

S₃ = 3a + 3d = 54
$\Rightarrow$ a + d = 18
S₂₀ = 10(2a + 19d)
$\Rightarrow$ 10(36 + 17d)
$\Rightarrow$ 1600 < 10(36 + 17d) < 1800
$\Rightarrow$ 160 < 36 + 17d < 180
$\Rightarrow$ 124 < 17d < 144
$\Rightarrow 7\frac{5}{17}<\mathrm{d}<8\frac{8}{17}$
Common difference will be natural number
$\Rightarrow$ d = 8 $\Rightarrow$ a = 10
$\Rightarrow$ a₁₁ = 10 + 10 × 8 = 90