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 :

Let the first term be $$a $$, and the common difference be $$d$$.

Sum of first three terms,  $$ S_3 = a + (a+d) + (a+2d) = 3a + 3d = 54 \Rightarrow a + d = 18 \quad ...(1) $$

Now, Sum of first 20 terms,  $$ S_{20} = \frac{20}{2}[2a + 19d] = 10(2a + 19d) $$

or, $$ S_{20} = 10{ 2(18-d) + 19d} = 360 + 170d $$  [Using (1) ]

$$S_{20}$$ lies between 1600 and 1800, so $$1600 < 360 + 170d < 1800 $$ 

$$\Rightarrow 1240 < 170d < 1440 $$

$$\Rightarrow \dfrac{1240}{170} < d < \dfrac{1440}{170} $$

$$ \Rightarrow 7.29 < d < 8.47 $$

Since The AP is made of positive integers so, $$d$$ is an integer, $$ d = 8 $$

And,$$ a = 18 - 8 = 10 $$

So, $$ a_{11} = a + 10d = 10 + 80 = 90 $$