Let a_1, a_2, a_3, ldots be in an A.P. such t | vedclass

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(A) Let the first term be $a_1 = a$ and the common difference be $d$.The sum of the first $12$ odd-indexed terms is given by $\sum_{k=1}^{12} a_{2k-1}

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$11$

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(A) Let the first term be $a_1 = a$ and the common difference be $d$.The sum of the first $12$ odd-indexed terms is given by $\sum_{k=1}^{12} a_{2k-1} = a_1 + a_3 + \ldots + a_{23} = -\frac{72}{5} a$.This is an $A.P.$ with $12$ terms,first term $a$,and common difference $2d$.The sum is $\frac{12}{2} [2a + (12-1)(2d)] = 6(2a + 22d) = 12a + 132d$.Equating this to the given value: $12a + 132d = -\frac{72}{5} a$.Multiplying by $5$: $60a + 660d = -72a$,which simplifies to $132a = -660d$,so $a = -5d$.We are given $\sum_{k=1}^{n} a_k = 0$,which is $\frac{n}{2} [2a + (n-1)d] = 0$.Since $n 
eq 0$,we have $2a + (n-1)d = 0$.Substituting $a = -5d$: $2(-5d) + (n-1)d = 0$.$-10d + nd - d = 0 \Rightarrow (n-11)d = 0$.Since $a_1 
eq 0$,$d 
eq 0$,therefore $n - 11 = 0$,which gives $n = 11$.

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

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