JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Sequences and Series

Let the A.P. have $2n$ terms, first term $a$ and common difference $d$. Each even-placed term exceeds the odd-placed term before it by $d$, and there are $n$ such pairs, so (sum of even terms) minus (sum of odd terms) is $nd=30-24=6$. The last term exceeds the first by $(2n-1)d=\frac{21}{2}$; since $(2n-1)d=2nd-d=12-d=\frac{21}{2}$, we get $d=\frac{3}{2}$ and then $n=\frac{6}{d}=4$, i.e. $8$ terms in all. The odd-placed terms form an A.P. with common difference $2d$, so $\frac{n}{2}[2a+(n-1)(2d)]=2[2a+9]=24$ gives $a=\frac{3}{2}$. The eight terms are $\frac{3}{2},3,\frac{9}{2},6,\frac{15}{2},9,\frac{21}{2},12$, whose integer members are $3,6,9,12$. Hence the number of integer terms is $4$.