JEE Main 2025 Jan 29 Shift 1, Mathematics Q13: Arithmetic, Geometric And Harmonic Progressions

JEE Main2025Jan 29, Shift 1Mathematics

Consider an A.P. of positive integers, whose sum of the first three terms is $54$ and the sum of the first twenty terms lies between $1600$ and $1800.$ Then its $1 1^{th}$ term is :

A
$84$
B
$122$
C
$90$
D
$108$

Solution

Let the first term be $a$ and the common difference $d .$ Then $S_{3} = 3 a + 3 d = 54 \Rightarrow a + d = 18.$

$S_{20} = 10 (2 a + 19 d) = 10 [2 (18 - d) + 19 d] = 10 (36 + 17 d) .$

The constraint $1600 < 10 (36 + 17 d) < 1800$ simplifies to $124 < 17 d < 144$ , i.e. $7.29 < d < 8.47.$

For an A.P. of positive integers $d$ must be a positive integer, so $d = 8$ and $a = 10.$

$a_{11} = a + 10 d = 10 + 80 = 90.$

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JEE Main 2025 Jan 29 Shift 1, Mathematics Q13: Arithmetic, Geometric And Harmonic Progressions

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