Question

Let $ a_1, a_2, a_3, ... $ 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

Hint:

Use the given conditions to form equations and solve for the first term and common ratio of the G.P.

Answer 1

The Correct Option is C

Let the \( 1^{st} \) term of G.P. be a and common ratio be r.
Given
\( a_3 a_5 = 729 \)
\( ar^2 \cdot ar^4 = 729 \)
\( a^2 r^6 = 729 \)
\( ar^3 = 27 \) ... (i)
Also, \( a_2 + a_4 = \frac{111}{4} \)
\( ar + ar^3 = \frac{111}{4} \)
\( ar = \frac{111}{4} - 27 = \frac{111 - 108}{4} = \frac{3}{4} \) ... (ii)
Dividing (i) by (ii):
\( \frac{ar^3}{ar} = \frac{27}{3/4} \)
\( r^2 = 36 \)
\( r = 6 \) (since the G.P. is increasing, r>0) From (ii):
\( a(6) = \frac{3}{4} \)
\( a = \frac{1}{8} \) Now, \( 24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) \)
\( = 24a(1 + r + r^2) \)
\( = 24 \cdot \frac{1}{8} (1 + 6 + 36) \)
\( = 3(43) \)
\( = 129 \)