Sequences & Series
GP and AP combined
nta_pyq_2023_jan
Grade 11
Question:
For two positive numbers $a, b$, if $a, b$ and $\dfrac{1}{18}$ are in a geometric progression, while $\dfrac{1}{a}$, $10$ and $\dfrac{1}{b}$ are in an arithmetic progression, then $16a + 12b$ is equal to ______.
Step-by-Step Solution
Key Concept: From GP: $\frac{a}{18} = b^2$, i.e., $a = 18b^2$. From AP: $\frac{1}{a} + \frac{1}{b} = 20$. Substitute and solve the cubic in $b$.
$a = 18b^2$ and $\frac{1}{a} + \frac{1}{b} = 20 \Rightarrow 360b^2 - 18b - 1 = 0 \Rightarrow b = \frac{1}{12}$, $a = \frac{1}{8}$. So $16a + 12b = 2 + 1 = 3$.
Correct Answer: 3