JEE PYQ: Sequence And Series Question 2
Question 2 - 2021 (16 Mar Shift 2)
Let $\frac{1}{16}$, $a$ and $b$ be in G.P. and $\frac{1}{a}$, $\frac{1}{b}$, $6$ be in A.P., where $a, b > 0$. Then $72(a + b)$ is equal to _______.
Show Answer
Answer: (14)
Solution
$a^2 = \frac{b}{16} \Rightarrow \frac{1}{b} = \frac{1}{16a^2}$
$\frac{2}{b} = \frac{1}{a} + 6$
$\frac{1}{8a^2} = \frac{1}{a} + 6$
$\frac{1}{a} - \frac{8}{a} - 48 = 0$
$\frac{1}{a} = 12, -4 \Rightarrow a = \frac{1}{12}, -\frac{1}{4}$
$a = \frac{1}{12}, a > 0$
$b = 16a^2 = \frac{1}{9}$
$\Rightarrow 72(a + b) = 6 + 8 = 14$
BETA
