Animated Solution for Mathematics - Sequence and Series: Let $a_1,a_2,a_3\dots$ be in an A.P. such that ${\sum }_{k=1}^{12}{a}_{2k-1}=-\frac{72}{5}{a}_1,a_1\ne 0$. If ${\sum }_{k=1}^n{a}_k=0$, then $n$ is:

We start with an Arithmetic Progression (A.P.) defined by its first term $a_1=a$ and a common difference $d$. The general term is given by $a_k=a+(k-1)d$.

The problem requires us to analyze the sum of the odd-indexed terms: $a_1+a_3+a_5+\cdots +a_{23}$.

These terms form a new sub-sequence. Because we skip every alternate term, the common difference of this new sequence is $D=2d$. We are dealing with $N=12$ such terms.

Using the standard sum formula $S_N=\frac{N}{2}[2a+(N-1)D]$, where $N=12$, the first term is $a$, and the common difference is $2d$, we calculate:

Simplifying this expression, we obtain:

This result represents the sum of our odd-indexed terms.

We now equate this sum to the condition provided in the problem: ${\sum }_{k=1}^{12}{a}_{2k-1}=-\frac{72}{5}{a}_1$. This creates a vital relationship between $a$ and $d$:

To simplify, we divide the entire equation by 12:

Multiplying by 5 to clear the fraction yields:

Rearranging the terms results in $11a+55d=0$, which simplifies to the golden relation:

Finally, we address the condition that the sum of the first $n$ terms of the original A.P. is zero. Using the sum formula $S_n=\frac{n}{2}[2a+(n-1)d]=0$:

Since $n eq 0$, we focus on the bracketed term: $2a+(n-1)d=0$. Substituting our relation $a=-5d$ into this equation gives:

Given that $a eq 0$ implies $d eq 0$, we must have $n-11=0$. Thus, the value of $n$ is 11.