In an arithmetic progression, if $$S_{40}=1030$$  and $$S_{12}=57$$, then $$S_{30}-S_{10} $$ is equal to:

We have, $$S_{40}=1030$$ and $$S_{12}=57$$

We know, $$ S_n = \frac{n}{2}[2a + (n-1)d] $$

So, $$ S_{40} = 20(2a + 39d) = 1030 \Rightarrow 2a + 39d = \dfrac{1030}{20} = 51.5 \quad ....(1) $$

And, $$S_{12} = 6(2a + 11d) = 57 \Rightarrow 2a + 11d = \dfrac{57}{6} = 9.5 \quad ....(2) $$

Subtracting , $$ (2a + 39d) - (2a + 11d) = 51.5 - 9.5 $$
$$\Rightarrow 28d = 42$$
$$\Rightarrow d = 1.5$$

Using this in eqn (2), $$ 2a + 11(1.5) = 9.5 $$
$$ \Rightarrow 2a + 16.5 = 9.5 $$
$$ \Rightarrow 2a = -7 \Rightarrow a = -3.5$$

Hence, $$ S_{30} - S_{10} = \text{sum of terms from } 11 \text{ to } 30 = \dfrac{20}{2}[a_{11} + a_{30}] $$

$$ a_{11} = a + 10d = -3.5 + 15 = 11.5,\quad a_{30} = a + 29d = -3.5 + 43.5 = 40 $$

So, $$ S_{30} - S_{10} = 10(11.5 + 40) = 10(51.5) = 515$$