Let the A.P. have first term $a$, common difference $d$, and an even number of terms $n=2m$.

We are given:

1. Sum of odd-positioned terms $=24$
2. Sum of even-positioned terms $=30$
3. Last term exceeds first term by $\dfrac{21}{2}$

We need the number of terms which are integers.

1. Write the odd and even terms separately

The A.P. is: $a,\ a+d,\ a+2d,\ a+3d,\dots, a+(2m-1)d$

Odd-positioned terms

These are: $a,\ a+2d,\ a+4d,\dots, a+(2m-2)d$ This is an A.P. with $m$ terms, first term $a$, common difference $2d$.

So, $S_{\text{odd}}=\frac{m}{2}\left[2a+(m-1)2d\right]=m(a+(m-1)d)$ Given: $m(a+(m-1)d)=24 \quad ...(1)$

Even-positioned terms

These are: $a+d,\ a+3d,\ a+5d,\dots, a+(2m-1)d$ This is also an A.P. with $m$ terms, first term $a+d$, common difference $2d$.

So, $S_{\text{even}}=\frac{m}{2}\left[2(a+d)+(m-1)2d\right]=m(a+md)$ Given: $m(a+md)=30 \quad ...(2)$

2. Use the difference of the two sums

Subtract (1) from (2): $m(a+md)-m(a+(m-1)d)=30-24$ $m\big(md-(m-1)d\big)=6$ $md=6$ So, $d=\frac{6}{m} \quad ...(3)$

3. Use the condition on first and last term

Last term $=a+(2m-1)d$. Given that last term exceeds first term by $\dfrac{21}{2}$: $a+(2m-1)d-a=\frac{21}{2}$ $(2m-1)d=\frac{21}{2}$ Using $d=\dfrac{6}{m}$: $(2m-1)\frac{6}{m}=\frac{21}{2}$ Multiply by $2m$: $12(2m-1)=21m$ $24m-12=21m$ $3m=12$ $m=4$ Hence, $n=2m=8$

So the A.P. has $8$ terms.

4. Find $a$ and $d$

From (3): $d=\frac{6}{4}=\frac{3}{2}$

From (1): $m(a+(m-1)d)=24$ $4(a+3d)=24$ $a+3d=6$ $a+3\cdot \frac{3}{2}=6$ $a+\frac{9}{2}=6$ $a=\frac{3}{2}$

Thus the A.P. is: $\frac{3}{2},\ 3,\ \frac{9}{2},\ 6,\ \frac{15}{2},\ 9,\ \frac{21}{2},\ 12$

5. Count the integer terms

The terms are: $\frac{3}{2},\ 3,\ \frac{9}{2},\ 6,\ \frac{15}{2},\ 9,\ \frac{21}{2},\ 12$

Integers among these are: $3,\ 6,\ 9,\ 12$

So the number of integer terms is: $4$

6. Check options

• A: $6$ ❌
• B: $4$ ✅
• C: $8$ ❌
• D: $10$ ❌

Therefore, the correct option is B.