Sequences and Series – Arithmetic Progression – JEE Main 24 January 2025 Shift 2

Solution:

Let $a$ be the first term and $d$ be the common difference.
Using the formula $S_n = \frac{n}{2}[2a + (n-1)d]$:

For $n=40$:
$$ 1030 = \frac{40}{2}[2a + 39d] \Rightarrow 1030 = 20(2a + 39d) $$
$$ 103 = 4a + 78d \quad \dots(1) $$

For $n=12$:
$$ 57 = \frac{12}{2}[2a + 11d] \Rightarrow 57 = 6(2a + 11d) $$
$$ 19 = 2(2a + 11d) \Rightarrow 19 = 4a + 22d \quad \dots(2) $$

Subtracting (2) from (1):
$$ (4a + 78d) – (4a + 22d) = 103 – 19 $$
$$ 56d = 84 \Rightarrow d = \frac{84}{56} = \frac{3}{2} $$

Substituting $d$ back into (2):
$$ 4a + 22\left(\frac{3}{2}\right) = 19 \Rightarrow 4a + 33 = 19 $$
$$ 4a = -14 \Rightarrow a = -\frac{7}{2} $$

Now we calculate $S_{30} – S_{10}$:
$$ S_{30} = \frac{30}{2}[2a + 29d] = 15(2a + 29d) = 30a + 435d $$
$$ S_{10} = \frac{10}{2}[2a + 9d] = 5(2a + 9d) = 10a + 45d $$
$$ S_{30} – S_{10} = (30a – 10a) + (435d – 45d) = 20a + 390d $$

Substitute values of $a$ and $d$:
$$ 20\left(-\frac{7}{2}\right) + 390\left(\frac{3}{2}\right) $$
$$ -70 + 585 = 515 $$

Ans. (2)