JEE Main 2025 Jan 24 Shift 2, Mathematics Q9: Arithmetic, Geometric And Harmonic Progressions

JEE Main2025Jan 24, Shift 2Mathematics

In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57$ , then $S_{30} - S_{10}$ is equal to:

A
$510$
B
$515$
C
$525$
D
$505$

Solution

Using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ :

$S_{40} = 20 (2 a + 39 d) = 1030 \Rightarrow 2 a + 39 d = \frac{1030}{20} = 51.5 \dots (1)$

$S_{12} = 6 (2 a + 11 d) = 57 \Rightarrow 2 a + 11 d = 9.5 \dots (2)$

Subtracting (2) from (1): $28 d = 42 \Rightarrow d = \frac{3}{2}$ . Then $2 a = 9.5 - 11 \cdot \frac{3}{2} = 9.5 - 16.5 = - 7 \Rightarrow a = - \frac{7}{2}$ .

Now:

$S_{30} - S_{10} = 15 (2 a + 29 d) - 5 (2 a + 9 d) = 20 a + 390 d$

$= 20 (- \frac{7}{2}) + 390 \cdot \frac{3}{2} = - 70 + 585 = 515$

Correct option: (2)

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JEE Main 2025 Jan 24 Shift 2, Mathematics Q9: Arithmetic, Geometric And Harmonic Progressions

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