Sequences and Series – Arithmetic Progression – JEE Main 02 April 2025 Shift 2

Solution:

Let the A.P. have $n$ terms (where $n$ is even), first term $a$, and common difference $d$.

$$a_{2}+a_{4}+…+a_{n}=30$$
$$a_{1}+a_{3}+…+a_{n-1}=24$$

$$(a_{2}-a_{1})+(a_{4}-a_{3})+…+(a_{n}-a_{n-1})=30-24$$
$$d+d+…+d = 6$$
$$\frac{n}{2}d = 6 \Rightarrow nd = 12$$

$$a_{n}-a_{1} = (n-1)d = \frac{21}{2}$$
$$nd-d = \frac{21}{2}$$

$$12-d = \frac{21}{2}$$
$$d = 12-\frac{21}{2} = \frac{3}{2}$$

$$n\left(\frac{3}{2}\right) = 12 \Rightarrow n = 8$$

The sequence has $8$ terms, meaning there are $\frac{n}{2} = 4$ odd terms. The common difference of the sequence of odd terms is $2d = 3$.
$$S_{odd} = \frac{n}{2}[2a+(n-1)D]$$
$$\frac{4}{2}[2a+(4-1)(3)] = 24$$
$$2[2a+9] = 24$$
$$2a+9 = 12 \Rightarrow 2a = 3 \Rightarrow a = \frac{3}{2}$$

The terms of the A.P. are:
$$\frac{3}{2}, \left(\frac{3}{2}+\frac{3}{2}\right), \left(\frac{3}{2}+2\left(\frac{3}{2}\right)\right), …$$
$$\frac{3}{2}, 3, \frac{9}{2}, 6, \frac{15}{2}, 9, \frac{21}{2}, 12$$

The integer terms in the sequence are $3, 6, 9, 12$.

Ans. (1)