JEE Main 2025 Apr 2 Shift 2, Mathematics Q3: Arithmetic, Geometric And Harmonic Progressions

JEE Main2025Apr 2, Shift 2Mathematics

The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by $\frac{21}{2}$ . Then the number of terms which are integers in the A.P. is :

A
$4$
B
$10$
C
$6$
D
$8$

Solution

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

Sum of even-indexed terms $-$ sum of odd-indexed terms $= \frac{n}{2} d = 30 - 24 = 6 \Rightarrow n d = 12$ .

The last term exceeds the first by $(n - 1) d = \frac{21}{2}$ , so $n d - d = \frac{21}{2} \Rightarrow d = 12 - \frac{21}{2} = \frac{3}{2}$ and hence $n = 8$ .

Sum of odd-indexed terms (there are $4$ of them, with first term $a$ and common difference $2 d = 3$ ): $\frac{4}{2} [2 a + 3 (3)] = 24 \Rightarrow 2 a + 9 = 12 \Rightarrow a = \frac{3}{2}$ .

The eight terms are $\frac{3}{2}, 3, \frac{9}{2}, 6, \frac{15}{2}, 9, \frac{21}{2}, 12$ . The integer terms are $3, 6, 9, 12$ — four in total.

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JEE Main 2025 Apr 2 Shift 2, Mathematics Q3: Arithmetic, Geometric And Harmonic Progressions

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