3. Given five different real numbers, taking any two of them to find their sum results in ten sum numbers, where the smallest two sums are sequentially $32, 36, 37$, and the largest two sums are 48 and 51. Then the largest number among these five numbers is $\qquad$

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3. $\frac{55}{2}$.

Let the five numbers be $a, b, c, d, e (a<b<c<d<e)$. Then
$$
\begin{array}{l}
a+b=32, a+c=36, \\
c+e=48, d+e=51 .
\end{array}
$$

Below, we show that $b+c=37$.
Since $c-b=4, d-c=3, d-b=7$, we have $a+d=(a+b)+(d-b)=39$.
Thus, $b+c=37$.
Then $2a=(a+b)+(a+c)-(b+c)=31$. Therefore, $a=15.5, b=16.5, c=20.5, e=27.5$, $d=23.5$, so the largest number is 27.5.