JEE Main 2025 Jan 28 Shift 1, Mathematics Q11: Arithmetic, Geometric And Harmonic Progressions

JEE Main2025Jan 28, Shift 1Mathematics

Let $T_{r}$ be the $r$ -th term of an A.P. If for some $m$ , $T_{m} = \frac{1}{25}$ , $T_{25} = \frac{1}{20}$ and $20 r = 1 \sum 25 T_{r} = 13$ , then $5 m r = m \sum 2 m T_{r}$ is equal to :

A
$112$
B
$126$
C
$98$
D
$142$

Solution

Use $T_{r} = a + (r - 1) d$ and the sum formula.

$20 S_{25} = 13 \Rightarrow 20 \cdot \frac{25}{2} [2 a + 24 d] = 13 \Rightarrow 250 (2 a + 24 d) = 13$ .

Equivalently $S_{25} = \frac{25}{2} (2 a + 24 d) = \frac{13}{20}$ , so the average term $\frac{S _{25}}{25} = a + 12 d = \frac{13}{500}$ .

Also $T_{25} = a + 24 d = \frac{1}{20} = \frac{25}{500}$ . Subtracting: $12 d = \frac{12}{500} \Rightarrow d = \frac{1}{500}$ , and then $a = \frac{1}{500}$ .

From $T_{m} = a + (m - 1) d = \frac{1}{25}$ : $\frac{1}{500} + \frac{m - 1}{500} = \frac{20}{500}$ , giving $m = 20$ .

Now $r = m \sum 2 m T_{r} = r = 20 \sum 40 T_{r} = S_{40} - S_{19}$ .

$S_{40} = \frac{40}{2} (2 a + 39 d) = 20 \cdot \frac{41}{500} = \frac{820}{500}$ .
$S_{19} = \frac{19}{2} (2 a + 18 d) = \frac{19}{2} \cdot \frac{20}{500} = \frac{190}{500}$ .

$\sum_{r = 20}^{40} T_{r} = \frac{820 - 190}{500} = \frac{630}{500} = \frac{63}{50}$ .

$5 m \sum_{r = m}^{2 m} T_{r} = 100 \cdot \frac{63}{50} = 126$ .

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JEE Main 2025 Jan 28 Shift 1, Mathematics Q11: Arithmetic, Geometric And Harmonic Progressions

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