Question

For the two positive numbers $a, b$, if $a, b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{a}, 10$ and $\frac{1}{b}$ are in an arithmetic progression, then $16 a+12 b$ is equal to

Answer 1

The correct answer is 3.
a,b,181​→GP
18a​=b2....(i)
a1​,10,b1​→AP
a1​+b1​=20
⇒a+b=20ab, from eq. (i); we get
⇒18b2+b=360b3
⇒360b2−18b−1=0{∵b=0}
⇒b=72018±324+1440​​
⇒b=72018+1764​​{∵b>0}
⇒b=121​
⇒a=18×1441​=81​
Now, 16a+12b=16×81​+12×121​=3