Sequences & Series
GP — Finding a₃+a₅+a₇
nta_pyq_2026_jan
Grade 11
Question:
Let $a_1,a_2,a_3,\ldots$ be a G.P. of increasing positive terms such that $a_2\cdot a_3\cdot a_4=64$ and $a_1+a_3+a_5=\dfrac{813}{7}$. Then $a_3+a_5+a_7$ is equal to:
Step-by-Step Solution
Key Concept: $a_3^3=a_2a_3a_4=64\Rightarrow a_3=4$. $a_1+a_5=813/7-4=785/7$ and $a_1a_5=a_3^2/1=16$. Solve for $a_1,a_5$.
$a_3+a_5+a_7=3252$.
Correct Answer: 3