The number of terms in an A.P. is even the sum of the odd terms in it is 24 and that the even terms is 30. If the last terms exceeds the first term by $10\frac{1}{2},$then the number of terms in the A.P. is

no of terms = 2n

$$\begin{gathered} {\mathrm{a}}_1+{\mathrm{a}}_3+{\mathrm{a}}_5 .... + {\mathrm{a}}_{2\mathrm{n}-1}=24 -\left(1\right) \\ {\mathrm{a}}_2+{\mathrm{a}}_4+{\mathrm{a}}_6 .... + {\mathrm{a}}_{2\mathrm{n}}=30 -\left(2\right) \\ \mathrm{and} {\mathrm{a}}_{2\mathrm{n}}={\mathrm{a}}_1+\frac{21}{2} \\ \mathrm{a}+(2\mathrm{n}-1)\mathrm{d}=\mathrm{a}+\frac{21}{2} \Rightarrow \mathrm{d}=\frac{21}{2(2\mathrm{n}-1)} \\ \mathrm{adding} \mathrm{eq} \left(1\right) \& \mathrm{eq} \left(2\right) \\ {\mathrm{S}}_{2\mathrm{n}}={\mathrm{S}}_4 \Rightarrow \frac{2\mathrm{n}}{2}\left[2\mathrm{a}+(2\mathrm{n}-1)\mathrm{d}\right]=54 \\ 2\mathrm{a}+\frac{21}{2}=\frac{54}{\mathrm{n}} \Rightarrow 2\mathrm{a}=\frac{54}{\mathrm{n}}-\frac{21}{2} \\ \mathrm{from} \mathrm{eq} \left(1\right) \\ \frac{\mathrm{n}}{2}\left[2\mathrm{a}+(\mathrm{n}-1)2\mathrm{d}\right]=24 \\ \Rightarrow \mathrm{n}=4 \Rightarrow \mathrm{no} \mathrm{of} \mathrm{terms}=8 \end{gathered}$$