Let 1/16,a and b be in G.P. and 1/a,1/( b),6 be in A.P., where a,b>0. Then 72(a+b) is equal to _______ .

Correct answer: 14. a^2=b/16⇒1/b=1/(16a^2) 2/b=1/a+6 1/(8a^2)=1/a+6 (1a^2-8)/a-48=0 1/a=12,-4⇒a=1/12,-1/4 a=1/12,a>0 b=16a^2=1/9 ⇒72(a+b)=6+8=14

Q56 / 90 · JEE Main 2021 (16 Mar Shift 2) · Mathematics

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 _______ .

Correct answer: 14

$a^{2} = \frac{b}{16} \Rightarrow \frac{1}{b} = \frac{1}{16 a^{2}}$

$\frac{2}{b} = \frac{1}{a} + 6$

$\frac{1}{8 a^{2}} = \frac{1}{a} + 6$

$\frac{1}{a^{2}} - \frac{8}{a} - 48 = 0$

$\frac{1}{a} = 12, - 4 \Rightarrow a = \frac{1}{12}, - \frac{1}{4}$

$a = \frac{1}{12}, a > 0$

$b = 16 a^{2} = \frac{1}{9}$

$\Rightarrow 72 (a + b) = 6 + 8 = 14$

Hence, the correct answer is 14.