Let $a,d,$ nad $2n $ be the frist term, common difference and total number of terms of an A.P. respectively i.e. $a + \left( {a + d} \right) + \left( {a + 2d} \right) + … + \left( {a + \left( {2n – 1} \right)d} \right)$

No. of even terms $=n$, of odd terms $=n$

Sum of odd terms:

${S_o} = \frac{n}{2}\left[ {2a + \left( {n – 1} \right)\left( {2d} \right)} \right] = 24\,\,\,\,$

$ \Rightarrow n\left[ {a + \left( {n – 1} \right)d} \right] = 24\,\,\,\,\,\,\,\,\,\,\,\,…..\left( i \right)$

Sum of even terms :

${S_e} = \frac{n}{2}\left[ {2\left( {a + d} \right) + \left( {n – 1} \right)2d} \right] = 30$

$ \Rightarrow n\left[ {a + d + \left( {n – 1} \right)d} \right] = 30\,\,\,\,\,\,\,\,\,\,\,\,…..\left( {ii} \right)$

Subtracting equation $(i)$ from $(ii)$ , we get $nd=6$      ….$(iii)$

Also, given that last term exceeds the first term by $\frac{{21}}{2}$

$a + \left( {2n – 1} \right)d = a + \frac{{21}}{2}$

$2nd – d = \frac{{21}}{2}$

$ \Rightarrow 2 \times 6 – \frac{{21}}{2} = d$       $(\,\,nd = 6)$

$d = \frac{3}{2}$

Putting value of $d$ in equation $(iii)$

$n = \frac{{6 \times 2}}{3} = 4$

Total no, of terms $ = 2n = 2 \times 4 = 8$