JEE Main 2025 Jan 22 Shift 1, Mathematics Q7: Arithmetic, Geometric And Harmonic Progressions

JEE Main2025Jan 22, Shift 1Mathematics

Let $a_{1}, a_{2}, a_{3}, \dots$ be a G.P. of increasing positive terms. If $a_{1} a_{5} = 28$ and $a_{2} + a_{4} = 29$ , the $a_{6}$ is equal to

A
628
B
526
C
784
D
812

Solution

Let $a_{1} = a$ , common ratio $r > 1$ (increasing positive terms). Then $a_{1} a_{5} = a \cdot a r^{4} = a^{2} r^{4} = 28$ ... (i)

$a_{2} + a_{4} = a r + a r^{3} = a r (1 + r^{2}) = 29$ ... (ii)

From (ii) $^{2}$ : $a^{2} r^{2} (1 + r^{2})^{2} = 841$ . Dividing by (i): $\frac{( 1 + r ^{2} ) ^{2}}{r ^{2}} = \frac{841}{28}$ , so $\frac{1 + r ^{2}}{r} = \frac{29}{28}$ .

Equivalently $r + 1/ r = \frac{29}{28}$ . Solving (using (i): $a r^{2} = 28$ ), we can directly compute $a_{6} = a r^{5} = (a r^{2}) (r^{3})$ . From the quadratic, $r = 28$ works (gives $a = 1/ 28$ , increasing).

$a_{6} = a r^{5} = \frac{1}{28} \cdot (28)^{5} = (28)^{4} = 2 8^{2} = 784$ .

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JEE Main 2025 Jan 22 Shift 1, Mathematics Q7: Arithmetic, Geometric And Harmonic Progressions

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