JEE Main 2025 Apr 3 Shift 1, Mathematics Q16: Arithmetic, Geometric And Harmonic Progressions

JEE Main2025Apr 3, Shift 1Mathematics

Let $a_{1}, a_{2}, a_{3}, \dots$ be a G.P. of increasing positive numbers. If $a_{3} a_{5} = 729$ and $a_{2} + a_{4} = \frac{111}{4}$ , then $24 (a_{1} + a_{2} + a_{3})$ is equal to

A
131
B
130
C
129
D
128

Solution

Write $a_{n} = a r^{n - 1}$ with $r > 1$ (increasing positive).

$a_{3} a_{5} = a^{2} r^{6} = (a r^{3})^{2} = 729 \Rightarrow a r^{3} = 27.$

$a_{2} + a_{4} = a r + a r^{3} = a r (1 + r^{2}) = \frac{111}{4} .$

Divide: $\frac{a r ^{3}}{a r} = r^{2} = \frac{27}{a r} .$ Also from the sum, $a r = \frac{111/4}{1 + r ^{2}} .$

Alternatively, use $a r (1 + r^{2}) = \frac{111}{4}$ and $a r \cdot r^{2} = 27$ . Subtracting in product form: let $u = a r$ , then $u (1 + r^{2}) = \frac{111}{4}$ and $u r^{2} = 27$ , so $u = \frac{111}{4} - 27 = \frac{111 - 108}{4} = \frac{3}{4} .$

Then $r^{2} = \frac{27}{u} = \frac{27 \cdot 4}{3} = 36 \Rightarrow r = 6, a = \frac{u}{r} = \frac{3/4}{6} = \frac{1}{8} .$

$24 (a + a r + a r^{2}) = 24 a (1 + r + r^{2}) = 24 \cdot \frac{1}{8} (1 + 6 + 36) = 3 \cdot 43 = 129.$

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JEE Main 2025 Apr 3 Shift 1, Mathematics Q16: Arithmetic, Geometric And Harmonic Progressions

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