JEE Mathematics 2025 Question with Solution

Given: $S_{40} = 1030$ and $S_{12} = 57$ for an arithmetic progression.

Find: $S_{30} - S_{10}$.

Use the sum formula of an arithmetic progression:

$$S_n = \frac{n}{2}\left(2a + (n-1)d\right)$$

From $S_{40} = 1030$,

$$1030 = \frac{40}{2}(2a + 39d) = 20(2a + 39d)$$

so,

$$2a + 39d = \frac{1030}{20} = 51.5$$

From $S_{12} = 57$,

$$57 = \frac{12}{2}(2a + 11d) = 6(2a + 11d)$$

so,

$$2a + 11d = \frac{57}{6} = 9.5$$

Subtracting,

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

Substitute into $2a + 11d = 9.5$:

$$2a + 16.5 = 9.5$$ $$2a = -7$$ $$a = -3.5$$

Now,

$$S_{30} = \frac{30}{2}(2a + 29d) = 15(2a + 29d)$$ $$2a + 29d = -7 + 43.5 = 36.5$$ $$S_{30} = 15 \times 36.5 = 547.5$$

Also,

$$S_{10} = \frac{10}{2}(2a + 9d) = 5(2a + 9d)$$ $$2a + 9d = -7 + 13.5 = 6.5$$ $$S_{10} = 5 \times 6.5 = 32.5$$

Therefore,

$$S_{30} - S_{10} = 547.5 - 32.5 = 515$$

The correct option is C. The solution also notes a discrepancy with the listed answer key, but the worked calculation gives $515$.

Given: $S_{40} = 1030$ and $S_{12} = 57$.

Find: $S_{30} - S_{10}$.

First find $a$ and $d$ from the given sum equations:

$$2a + 39d = 51.5$$ $$2a + 11d = 9.5$$

Hence,

$$d = 1.5, \qquad a = -3.5$$

Now interpret

$$S_{30} - S_{10}$$

as the sum of terms from the $11^{\text{th}}$ to the $30^{\text{th}}$ term. Using the sum formula values obtained above,

$$S_{30} = 547.5$$

and

$$S_{10} = 32.5$$

So,

$$S_{30} - S_{10} = 515$$

Thus, the required value is $515$, so the correct option is C.