Let $a_1, a_2, a_3 \ldots$ be in an A.P. such that $\sum_{\mathrm{k}=1}^{12} \mathrm{a}_{2 \mathrm{k}-1}=-\frac{72}{5} \mathrm{a}_1, \mathrm{a}_1
eq 0$. If $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{k}}=0$, then n is:

Question

Let $a_1, a_2, a_3 \ldots$ be in an A.P. such that $\sum_{\mathrm{k}=1}^{12} \mathrm{a}_{2 \mathrm{k}-1}=-\frac{72}{5} \mathrm{a}_1, \mathrm{a}_1 
eq 0$. If $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{k}}=0$, then n is:

Aicharya · Accepted Answer

Let $a_1=a$, common difference $=d$
$a_1+a_3+a_5+\ldots \ldots+a_{23}=-\frac{72}{5} a$
$\frac{12}{2}[2 a+11 \times 2 d]=-\frac{72}{5} a$
$12 a+132 d=-\frac{72}{5} a$
$132 a+132 \times 5 d=0$
$\mathrm{a}=-5 \mathrm{~d}$
$\frac{\mathrm{n}}{2}(2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d})=0 \Rightarrow-10 \mathrm{~d}+\mathrm{nd}-\mathrm{d}=0$
$\mathrm{n}=11$ *

Let $a_1, a_2, a_3 \ldots$ be in an A.P. such that $\sum_{\mathrm{k}=1}^{12} \mathrm{a}_{2…

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