JEE Mathematics 2024 Question with Solution

Given: $S_{20} = 790$ and $S_{10} = 145$ for an arithmetic progression with first term $a$ and common difference $d$.

Find: $S_{15} - S_{5}$.

Use the sum formula of an AP:

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

From $S_{20} = 790$,

$$\frac{20}{2}\left[2a + 19d\right] = 790$$ $$10\left[2a + 19d\right] = 790$$ $$2a + 19d = 79$$

From $S_{10} = 145$,

$$\frac{10}{2}\left[2a + 9d\right] = 145$$ $$5\left[2a + 9d\right] = 145$$ $$2a + 9d = 29$$

Subtracting,

$$(2a + 19d) - (2a + 9d) = 79 - 29$$ $$10d = 50$$ $$d = 5$$

Substitute into $2a + 9d = 29$:

$$2a + 9 \times 5 = 29$$ $$2a + 45 = 29$$ $$2a = -16$$ $$a = -8$$

Now,

$$S_{15} = \frac{15}{2}\left[2(-8) + 14 \times 5\right]$$ $$S_{15} = \frac{15}{2}(-16 + 70)$$ $$S_{15} = \frac{15}{2} \times 54 = 405$$

Also,

$$S_{5} = \frac{5}{2}\left[2(-8) + 4 \times 5\right]$$ $$S_{5} = \frac{5}{2}(-16 + 20)$$ $$S_{5} = \frac{5}{2} \times 4 = 10$$

Therefore,

$$S_{15} - S_{5} = 405 - 10 = 395$$

The correct option is A. The solution also shows a discrepancy where one place labels the correct option as B, but the worked calculation gives $395$, which matches option A.

Given: $S_{20} = 790$ and $S_{10} = 145$.

Find: $S_{15} - S_{5}$.

Observe that $S_{15} - S_{5}$ is the sum of the $6$th to $15$th terms, while $S_{20} - S_{10}$ is the sum of the $11$th to $20$th terms. In an AP, both are sums of $10$ consecutive terms, so their averages differ by exactly $5d$. The given working first finds $d = 5$.

From the equations,

$$2a + 19d = 79$$

and

$$2a + 9d = 29$$

Subtracting gives

$$d = 5$$

Then evaluate the required expression using the AP sum formula, which gives

$$S_{15} - S_{5} = 395$$

Therefore, the correct option is A.