Sequence and Series – Arithmetic Progression – JEE Main 2 Apr 2025 Shift 1

Solution:

Let the first term be $a$ and the common difference be $d$.
The first sum is of terms $a_1, a_3, a_5, \dots, a_{23}$.
This series has 12 terms with first term $a$ and common difference $2d$.

Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)D]$:
$$ \frac{12}{2}[2a + (12-1)(2d)] = -\frac{72}{5}a $$

Simplify the equation:
$$ 6[2a + 22d] = -\frac{72}{5}a $$
$$ 2a + 22d = -\frac{12}{5}a $$
$$ 10a + 110d = -12a $$
$$ 22a = -110d \Rightarrow a = -5d $$

Now, consider the second condition $\sum_{k=1}^{n}a_{k}=0$.
$$ \frac{n}{2}[2a + (n-1)d] = 0 $$

Since $n \neq 0$, we have:
$$ 2a + (n-1)d = 0 $$

Substitute $a = -5d$:
$$ 2(-5d) + (n-1)d = 0 $$
$$ -10d + nd – d = 0 $$
$$ d(n – 11) = 0 $$

Since $a \neq 0$, implies $d \neq 0$, so:
$$ n – 11 = 0 \Rightarrow n = 11 $$

Ans. (1)