JEE Main 2021 (Online) 16th March Evening ShiftQuestion 1
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 $\_\_\_\_\_\_$ .
Sequence and Series · Mixed Progressions (AP-GP)
Answer$14$
Solution
Formulas used. (1) $x,y,z$ in G.P. $\Rightarrow y^2=xz$(2) $x,y,z$ in A.P. $\Rightarrow 2y=x+z$Step 1 - G.P. $a^2=\dfrac{b}{16}$, so $b=16a^2$. Step 2 - A.P. $$\frac2b=\frac1a+6\ \Longrightarrow\ \frac1{8a^2}=\frac1a+6\ \Longrightarrow\ 48a^2+8a-1=0 .$$ $(12a-1)(4a+1)=0$ and $a>0$, so $a=\dfrac1{12}$ and $b=\dfrac{16}{144}=\dfrac19$. Step 3. $$72(a+b)=6+8=14 .$$ Answer: 14