Let the A.P. have first term $a$ and common difference $d$.

The number of terms is $2k$.

So the terms are: $a,\, a+d,\, a+2d,\, \dots,\, a+(2k-1)d$

1. Sum of odd-positioned terms

Odd-positioned terms are: $a,\ a+2d,\ a+4d,\ \dots,\ a+(2k-2)d$ This is an A.P. with:

• first term $a$
• common difference $2d$
• number of terms $k$

Hence, $S_{\text{odd}}=\frac{k}{2}\big[a+\{a+(2k-2)d\}\big]$ Given this sum is $40$: $\frac{k}{2}\big(2a+(2k-2)d\big)=40$ $k\big(a+(k-1)d\big)=40 \qquad (1)$

2. Sum of even-positioned terms

Even-positioned terms are: $a+d,\ a+3d,\ a+5d,\ \dots,\ a+(2k-1)d$ This is also an A.P. with:

• first term $a+d$
• last term $a+(2k-1)d$
• number of terms $k$

Hence, $S_{\text{even}}=\frac{k}{2}\big[(a+d)+(a+(2k-1)d)\big]$ Given this sum is $55$: $\frac{k}{2}\big(2a+2kd\big)=55$ $k(a+kd)=55 \qquad (2)$

3. Difference between last and first term

Last term $=$ $a+(2k-1)d$.

Given last term exceeds first term by $27$: $[a+(2k-1)d]-a=27$ $(2k-1)d=27 \qquad (3)$

4. Use equations (1) and (2)

Subtract (1) from (2): $k(a+kd)-k(a+(k-1)d)=55-40$ $k\big(kd-(k-1)d\big)=15$ $kd=15$ $d=\frac{15}{k} \qquad (4)$

Now substitute into (3): $(2k-1)\cdot \frac{15}{k}=27$ $15(2k-1)=27k$ $30k-15=27k$ $3k=15$ $k=5$

5. Check with options

The correct option is: $\boxed{5}$ So, option $\boxed{D}$ is correct.