Sequences and Series – Geometric Progression – JEE Main 03 April 2025 Shift 1

Solution:

$$a_n = ar^{n-1}$$

$$a_3 a_5 = 729$$

$$(ar^2)(ar^4) = 729$$

$$a^2 r^6 = 729$$

$$ar^3 = 27$$

$$a_2 + a_4 = \frac{111}{4}$$

$$ar + ar^3 = \frac{111}{4}$$

$$ar + 27 = \frac{111}{4}$$

$$ar = \frac{111}{4} – 27$$

$$ar = \frac{111 – 108}{4}$$

$$ar = \frac{3}{4}$$

$$\frac{ar^3}{ar} = \frac{27}{3/4}$$

$$r^2 = 36$$

$$r = 6$$

$$a(6) = \frac{3}{4}$$

$$a = \frac{1}{8}$$

$$24(a_1 + a_2 + a_3) = 24(a + ar + ar^2)$$

$$= 24a(1 + r + r^2)$$

$$= 24 \left(\frac{1}{8}\right) (1 + 6 + 36)$$

$$= 3(43)$$

$$= 129$$

Ans. (3)