Consider three sequences an A.P., a G.P. and a H.P. as follows
$a, A_{1}, A_{2}, A_{3} ....... A_{9}, b (A . P .)$
$a, G_{1}, G_{2}, G_{3}, .... G_{9}, b (G . P .);$
$a, H_{1}, H_{2}, H_{3}, ...... H_{9}, b (H . P .)$
If $a, b$ are both roots of the equation $x^{2} - (p + 1) x + 16 = 0, p \in R$ , then the value of $\prod_{r = 1}^{4} A_{r} H_{10 - r} G_{10 - 2 r}$ is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$4^{9}$

$4^{12}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\underset{}{\underset{⏟}{A_{1} H_{9} G_{8}}} \underset{}{\underset{⏟}{A_{2} H_{8} G_{6}}} \underset{}{\underset{⏟}{A_{3} H_{7} G_{4}}} \underset{}{\underset{⏟}{A_{4} H_{6} G_{2}}}$
Now $A_{1} H_{9} = A_{2} H_{8} = A_{3} H_{7} = a b = 16$
$∴$ Expression $= 16^{4} \times G_{2} G_{4} G_{6} G_{8}$
Now, $G_{2} G_{8} = G_{4} G_{6} = a b = 16$
$∴ 16^{4} \times 16^{2} = 4^{12}$