$\therefore \mathrm{b}^2=\mathrm{ac}$
Also
$\log _{\circ}\left(\frac{a}{2 b}\right), \log _{\circ}\left(\frac{2 b}{3 c}\right), \log _{\circ}\left(\frac{3 c}{a}\right)$ are in $A.P.$
$\left(\frac{2 b}{3 \mathrm{c}}\right)^2=\frac{\mathrm{a}}{2 \mathrm{~b}} \times \frac{3 \mathrm{c}}{\mathrm{a}} $
$ \frac{\mathrm{b}}{\mathrm{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$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$T_{\mathrm{I}}=V_{\mathrm{r}+1}-V_{\mathrm{I}}-2 \text { and } \mathrm{Q}_{\mathrm{I}}=T_{\mathrm{r}+1}-\mathrm{T}_{\mathrm{r}} \text { for } \mathrm{r}=1,2, \ldots$
$1.$ The sum $V_1+V_2+\ldots+V_n$ is
$(A)$ $\frac{1}{12} n(n+1)\left(3 n^2-n+1\right)$
$(B)$ $\frac{1}{12} n(n+1)\left(3 n^2+n+2\right)$
$(C)$ $\frac{1}{2} n\left(2 n^2-n+1\right)$
$(D)$ $\frac{1}{3}\left(2 n^3-2 n+3\right)$
$2.$ $\mathrm{T}_{\mathrm{T}}$ is always
$(A)$ an odd number $(B)$ an even number
$(C)$ a prime number $(D)$ a composite number
$3.$ Which one of the following is a correct statement?
$(A)$ $Q_1, Q_2, Q_3, \ldots$ are in $A.P.$ with common difference $5$
$(B)$ $\mathrm{Q}_1, \mathrm{Q}_2, \mathrm{Q}_3, \ldots$ are in $A.P.$ with common difference $6$
$(C)$ $\mathrm{Q}_1, \mathrm{Q}_2, \mathrm{Q}_3, \ldots$ are in $A.P.$ with common difference $11$
$(D)$ $Q_1=Q_2=Q_3=\ldots$
Give the answer question $1,2$ and $3.$