You're a bit vague about what you mean by "average" price. Do you mean the average price you paid for oranges or the average price oranges were transacted on the market. If the first, you account for the ones you sold as an income instead of a loss, so you should give it the opposite sign. But if you just are interested in the average price at which transactions occurred, not accounting for whether they were sold or bought, then the simple formula with absolute values is OK. What do you want?
Suppose you're interested in the average price you paid for the oranges.
You have used an amount of money equal to $2\times 5+3\times 6-3\times 4=16$ dollars to pay for oranges. You are left with two oranges, so you could say the average price per orange you paid is $8$ dollars.
On the other hand, if you're interested in the average transaction price, then you might need:
$$\frac{2\times 5+3\times 6+3\times 4}{2+3+3}=5 \text{ dollars.}$$
In your edited version, you say you want to evaluate "worth", I think averages are not really useful. Totals would be more useful. In fact, a good double-entry bookkeeping system that keeps track of your amount of oranges and your capital would be more useful here.
Perhaps the easiest way to see what’s going on is to look at an extreme case. Suppose that you start with just two initial values, $10$ and $1000$, and increase one of them by $10$% while leaving the other unchanged. The average of the two percentage increases is $\frac12(10+0)=5$% no matter which of the two numbers you increased. The percentage increase in the total, however, very clearly depends on which of the two was increased: increasing the $10$ by $10$% increases the total by $1$, from $1010$ to $1011$, but increasing the $1000$ by $10$% increases the total by $100$, from $1010$ to $1110$. Clearly the latter is a much larger percentage increase than the former; the actual figures are about $0.099$% in the first case and about $9.9$% in the second.
The point is that when you increase one of the items by a certain percentage, the effect on the total depends not only on the percentage, but also on the size of the item.
Edit: And that’s exactly why your weighted averages work: the individual percentage increases need to be weighted by the sizes of the original items. Let the original amounts be $a_1,\dots,a_n$, and let the percentage increases, expressed as fractions, be $p_1,\dots,p_n$. The amounts after the increase are $a_k(1+p_k)=a_k+a_kp_k$ for $k=1,\dots,n$, so the the total increase is $$a_1p_1+a_2p_2+\cdots+a_np_n\;.$$ Let $A=a_1+a_2+\cdots+a_n$ be the original total; then the fractional increase is
$$\frac{a_1p_1+a_2p_2+\cdots+a_np_n}A=\frac{a_1}Ap_1+\cdots+\frac{a_n}Ap_n\;,$$
which is indeed the weighted mean of the percentages when they are weighted by the relative sizes of the original amounts.
Best Answer
You have items $x_1, \ldots x_n$ with weights $w_1, \ldots, w_n$. The weighted average would be $$ \frac{\sum_{k=1}^n x_i w_i}{\sum_{k=1}^n w_i}. $$
In other words, let $W = \sum_{i=1}^n w_i$ and you can produce new weights $$ u_i = \frac{w_i}{W} $$ with the property that $$ \sum_{i=1}^n u_i = \sum_{i=1}^n \frac{w_i}{W} = \frac{1}{W} \sum_{i=1}^n w_i = 1 $$ and then compute $$ \sum_{i=1}^n u_i x_i = \sum_{i=1}^n \frac{w_i}{W} x_i = \frac{1}{W} \sum_{i=1}^n w_i x_i = \frac{\sum_{i=1}^n w_i x_i}{\sum_{i=1}^n w_i} $$ as noted above.