How do stoichiometric coefficients effect the half-life of a reaction?
Consider what happens to your argument if I rewrite the equation in the following equivalent form:
$$\ce{A -&gt; \frac{3}{2}B +\frac{1}{2}C +D}$$
That is, you write


  ... since rate of consumption of A is double the rate of reaction as can be seen from the stoichiometric coefficients


In contrast, the rate of the reaction, in this case, is defined to be the rate of decay of $\ce{A}$,
$$-\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = k[\ce{A}]$$
When you look at it like this, it is more clear that the stoichiometry should not have an effect on the decay process as you claimed.

Furthermore, if we integrate the equation above after separating by variables,
$$\int_{[\ce{A}_0]}^{[\ce{A_t}]}\frac{1}{[\ce{A}]}\mathrm{d}[\ce{A}] = \int_0^t-k\mathrm{d}t$$
we get,
$$\ln\left(\frac{[\ce{A_t}]}{[\ce{A}_0]}\right)=-kt.$$

If we try to find the time at which the original concentration has halved, we get the formula you have written above which is,
$$t_{\frac12}=\frac{\ln(2)}{k}.$$

The important point here, and this isn't true for other reaction-orders, is that the half-life is independent of initial concentration. This is not what one would intuitively expect, and indeed it's not true in general, but it is characteristic of all exponential-decay processes.

Indeed, this is a very deep result which doesn't tend to get emphasized enough in my opinion.

It is true that sometimes you can say that a reaction will be second order based on stoichiometric coefficients, but you are told this is first order, so that is probably the source of your confusion.

So, your textbook is right.