The number of terms of an A.P. is even; the sum of all the odd terms is 24 , the sum of all the even terms is 30 and the last term exceeds the first…

$\mathrm{a}_2+\mathrm{a}_4+\ldots+\mathrm{a}_{\mathrm{n}}=30$a2​+a4​+…+an​=30 ...(1)

$\mathrm{a}_1+\mathrm{a}_3+\ldots+\mathrm{a}_{\mathrm{n}-1}=24$a1​+a3​+…+an−1​=24 ...(2)

$$(1)-(2)$$(1)−(2)

$$\left(a_2-a_1\right)+\left(a_4-a_3\right) \ldots\left(a_n-a_{n-1}\right)=6$$(a2​−a1​)+(a4​−a3​)…(an​−an−1​)=6

$$\begin{aligned} & \Rightarrow \frac{\mathrm{n}}{2} \mathrm{~d}=6 \Rightarrow \mathrm{nd}=12 \\ & \mathrm{a}_{\mathrm{n}}-\mathrm{a}_1=(\mathrm{n}-1) \mathrm{d}=\frac{21}{2} \\ & \Rightarrow \mathrm{nd}-\mathrm{d}=\frac{21}{2} \Rightarrow 12-\frac{21}{2}=\mathrm{d} \\ & \Rightarrow \mathrm{d}=\frac{3}{2}, \mathrm{n}=8\end{aligned}$$​⇒2n​ d=6⇒nd=12an​−a1​=(n−1)d=221​⇒nd−d=221​⇒12−221​=d⇒d=23​,n=8​

$\begin{aligned}
& \text { Sum of odd terms }=\frac{4}{2}[2 \mathrm{a}+(4-1) 3]=24 \
& \Rightarrow \mathrm{a}=\frac{3}{2} \
& \text { A.P. } \Rightarrow \frac{3}{2}, 3, \frac{9}{2}, 6, \frac{15}{2}, 9, \frac{21}{2}, 12
\end{aligned}$

no. of integer terms $=4$=4