Question ID: #1128
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
- (1) 131
- (2) 130
- (3) 129
- (4) 128
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)
Was this solution helpful?
YesNo