How much do you want?

Let the number of term be $n = 2m$ , the first term be $a$ and the difference be $d$ .

Then, the sum of odd terms:

$\quad \quad \dfrac {m[2a+2d(m-1)]}{2} = am + dm^2 - dm = 24\quad ...(1)$

The sum of even terms:

$\quad \quad \dfrac {m[2a+2d+2d(m-1)]}{2} = am + dm^2 = 30\quad ...(2)$

Equation 2 - Equation 1: $\quad \Rightarrow dm = 6$

Also the last term exceeds the first by $10.5$ : $\quad \Rightarrow a + (2m-1)d - a = 10.5 \quad \Rightarrow 2dm - d = 10.5$

$\quad \Rightarrow 2(6) - d = 10.5 \quad \Rightarrow d = 1.5$

From $dm = 6 \quad \Rightarrow m = 4 \quad \Rightarrow n = \boxed {8}$

$Let\quad total\quad terms\quad n\quad \& \quad difference\quad is\quad d.\\ \\ { S }_{ even }{ -S }_{ odd }=\sum _{ m=1 }^{ { n }/{ 2 } }{ \left( { a }_{ 2m }-{ a }_{ 2m-1 } \right) } =\sum _{ m=1 }^{ { n }/{ 2 } }{ d= } \cfrac { n }{ 2 } d\\ \\ \cfrac { n }{ 2 } d=30-24=6\quad \Rightarrow nd=12\\ \\ \& \quad \left( { a }_{ n }{ -a }_{ 1 } \right) =\left( n-1 \right) d=10.5=\cfrac { 21 }{ 2 } \\ \\ \Rightarrow 1-\cfrac { 1 }{ n } =\cfrac { \left( n-1 \right) d }{ nd } =\cfrac { 21 }{ 24 } =1-\cfrac { 1 }{ 8 } \\ \\ \Rightarrow n=8$

A= 1.5 d= 1.5