Sequences And Series Question 10
Question 10 - 2024 (29 Jan Shift 2)
If $\log _e a, \log _e b, \log _e c$ are in an A.P. and $\log _e a-\log _e 2 b, \log _e 2 b-\log _e 3 c, \log _e 3 c-\log _e a$ are also in an A.P, then $a: b: c$ is equal to
(1) $9: 6: 4$
(2) $16: 4: 1$
(3) $25: 10: 4$
(4) $6: 3: 2$
Show Answer
Answer (1)
Solution
$\log _e a, \log _e b, \log _e c$ are in A.P.
$\therefore b^{2}=ac \ldots(i)$
Also
$\log _e\left(\frac{a}{2 b}\right), \log _e\left(\frac{2 b}{3 c}\right), \log _e\left(\frac{3 c}{a}\right)$ are in A.P.
$\left(\frac{2 b}{3 c}\right)^{2}=\frac{a}{2 b} \times \frac{3 c}{a}$
$\frac{b}{c}=\frac{3}{2}$
Putting in eq. (i) $b^{2}=a \times \frac{2 b}{3}$
$\frac{a}{b}=\frac{3}{2}$
$a: b: c=9: 6: 4$
