JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Sequences and Series

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

, then

a_{6}

is equal to

Answer: (C)

Solution

Let the G.P. have first term

a_{1}

and common ratio

r > 0

(increasing positive terms).

Note

a_{1} a_{5} = a_{3}^{2}

, since

a_{3} = a_{1} r^{2}

gives

a_{3}^{2} = a_{1}^{2} r^{4} = a_{1} (a_{1} r^{4}) = a_{1} a_{5}

. So

a_{3}^{2} = 28

.

Also

a_{2} a_{4} = (a_{1} r) (a_{1} r^{3}) = a_{1}^{2} r^{4} = a_{1} a_{5} = 28

, so

a_{2}

and

a_{4}

are the two roots of

t^{2} - (a_{2} + a_{4}) t + a_{2} a_{4} = 0 ⟹ t^{2} - 29 t + 28 = 0 ⟹ (t - 1) (t - 28) = 0.

{a_{2}, a_{4}} = {1, 28}

. Since the terms are increasing,

a_{2} < a_{4}

, so

a_{2} = 1, a_{4} = 28

.

Then

r^{2} = \frac{a _{4}}{a _{2}} = 28

, and

a_{6} = a_{4} \cdot r^{2} = 28 \times 28 = 784.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option