在 $(3+2ax)^5$ 的展开式中，$x^3$ 的系数是在 $(2+ax)^6$ 的展开式中 $x^2$ 系数的六倍。

求常数 $a$ 的值。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

先写出两项的系数，再把等式移到一边并提取公因式。保留公因式可能给出的零解。

The coefficient of $x^3$ in $(3+2ax)^5$ is

The coefficient of $x^2$ in $(2+ax)^6$ is $\binom62 2^4a^2=240a^2$. Thus

The nonzero solution is $a=2$. The value $a=0$ also satisfies the stated coefficient equality, as both coefficients are zero then.