Question

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

Answer 1

11

Answer 2

Let the arithmetic progression (AP) be denoted by $a_1, a_2, a_3, \dots$ with the first term $a_1$ and common difference $d$.

The terms $a_{2k-1}$ for $k=1, 2, \dots, 12$ are $a_1, a_3, a_5, \dots, a_{23}$. This sequence of terms is itself an AP with the first term $A = a_1$ and a common difference $D = a_3 - a_1 = (a_1 + 2d) - a_1 = 2d$. There are $N=12$ terms in this sum.

The sum of these terms is given by the formula for the sum of an AP: $S_N = \frac{N}{2}[2A + (N-1)D]$ $\sum_{k=1}^{12} a_{2k-1} = \frac{12}{2}[2a_1 + (12-1)(2d)]$ $= 6[2a_1 + 11(2d)]$ $= 6[2a_1 + 22d]$ $= 12a_1 + 132d$.

We are given that $\sum_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1$. Therefore, $12a_1 + 132d = -\frac{72}{5} a_1$.

Rearranging the terms to find a relationship between $a_1$ and $d$: $132d = -\frac{72}{5} a_1 - 12a_1$ $132d = \left(-\frac{72}{5} - \frac{60}{5}\right) a_1$ $132d = -\frac{132}{5} a_1$.

Since $a_1 \neq 0$, we can divide by $132a_1$: $d = -\frac{1}{5} a_1$. This implies $a_1 = -5d$. Since $a_1 \neq 0$, it follows that $d \neq 0$.

The second condition is $\sum_{k=1}^{n} a_k = 0$. This is the sum of the first $n$ terms of the AP, $S_n$. The formula for $S_n$ is: $S_n = \frac{n}{2}[2a_1 + (n-1)d]$.

We are given $S_n = 0$. $\frac{n}{2}[2a_1 + (n-1)d] = 0$. Since $n$ represents the number of terms, $n$ must be a positive integer, so $n \neq 0$. Thus, we must have: $2a_1 + (n-1)d = 0$.

Substitute the relation $a_1 = -5d$ into this equation: $2(-5d) + (n-1)d = 0$ $-10d + (n-1)d = 0$.

Factor out $d$: $d(-10 + n-1) = 0$ $d(n-11) = 0$.

Since we established that $d \neq 0$ (because $a_1 \neq 0$), we can conclude that: $n-11 = 0$ $n = 11$.