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I found the following Problem on page 76 :
"C : 9. Let $a_1,\; a_2,\,\dots\;,a_{10}$ be real numbers such that $a_{10}\neq 0$ and
$\frac{a_1}{a_2+a_3+\dots +a_{10}}=\frac{a_1+a_2}{a_3+a_4+\dots +a_{10}}=\dots=\frac{a_1+a_2+\dots +a_9}{a_{10}}=K.$
Show that $K\in \left \{-1,\;\frac{a_1}{a_{10}} \right \}$. Generalization.
{ Author : } Mihai DICU, teacher, Craiova"
ANSWER CiP
The problem's assumptions are insufficient. In fact, the condition that other denominators appearing in the problem statement must also be non-zero is missing. We address this in the following
GENERALIZATION
Let $m\geqslant 3$ and real numbers $a_1,\;a_2,\;\dots \;,\;a_m$ sucht that
$a_m \neq 0,\;a_{m-1}+a_m \neq 0, \dots ,\;a_2+\dots +a_m \neq 0. \tag{I}$
If $K=\frac{a_1}{a_2+\dots +a_m}=\frac{a_1+a_2}{a_3+\dots +a_m}=\dots =\frac{a_1+\dots +a_{m-1}}{a_m}$ (E)
then $K\in \left \{-1,\;\frac{a_1}{a_m} \right \}.$
ANSWER CiP to GENERALIZATION
We have one of the following possibilities :
EITHER
$a_1+a_2+\dots a_m=0 \tag {$S_1$}$
when $K=-1$
OR
$a_2=\dots=a_{m-1}=0\;,\;\;a_1,\,a_m\;-\;being\;arbitrary\;, \;a_m\neq 0 \tag{$S_2$}$
when $K=\frac{a_1}{a_m}.$
Solution CiP
SUFFICIENCY
We fix an arbitrary index $p\in\{1,\;2,\dots ,m-1\}.$
$(S_1)\;\Rightarrow\; a_1+\dots +a_p=-(a_{p+1}+\dots +a_m)\underset{(I)}{\neq}0\;\Rightarrow\;\frac{a_1+\dots +a_p}{a_{p+1}+\dots +a_m}=-1$
so the equalities (E) are satisfied.
$(S_2)\;\Rightarrow\;\frac{a_1+\dots +a_p}{a_{p+1}+\dots +a_m}=\frac{a_1}{a_m}$
and equalities (E) are again satisfied.
NECESSITY
(to continue)
ciorna :
We fix an arbitrary index $p$. The equalities EE are satisfied. Alternatively, they are all zero, in which case the equalities EE are again satisfied.
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