Let $a_1,a_2,a_3,\dots$ be in an A.P. such that $\displaystyle\sum_{k=1}^{12}a_{2k-1}=-\tfrac{72}{5}a_1$, $a_1\neq0$. If $\displaystyle\sum_{k=1}^n a_k=0$, then $n$ is:

Step 1: Express the sum of odd terms

The given sum involves terms with odd indices: $a_1, a_3, \dots, a_{23}$. This is an arithmetic progression with 12 terms, where the first term is $a_1$ and the common difference is $2d$ (since we are skipping terms).

Step 2: Calculate the sum of the odd terms

Using the formula for the sum of an arithmetic progression, $S_m = \frac{m}{2}(2A + (m-1)D)$, where $m=12$, $A=a_1$, and $D=2d$. We substitute these values to find the sum of the odd terms.

Step 3: Formulate an equation for the common difference

We equate the calculated sum of odd terms to the given expression, $-\frac{72}{5}a_1$. This allows us to establish a relationship between the first term $a_1$ and the common difference $d$. We simplify the equation to find $a_1$ in terms of $d$.

Step 4: Set up the sum of the first n terms

The problem states that the sum of the first $n$ terms of the arithmetic progression is 0. We use the formula for the sum of an arithmetic progression, $S_n = \frac{n}{2}(2a_1 + (n-1)d)$, and set it equal to 0.

Step 5: Solve for n

Substitute $a_1 = -5d$ into the equation for the sum of the first $n$ terms. Since $a_1 \neq 0$, it implies $d \neq 0$. Also, $n$ cannot be 0 for a sum of terms. Therefore, the only possibility for the sum to be 0 is if the term $(n-11)$ is 0.

Step 6: Determine the value of n

From the previous step, we deduced that $n-11=0$. Solving for $n$ gives us the final answer.

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