==> ${1 \over {\ln a.\ln b.\ln c}}[{(\ln a)^3} + {(\ln b)^3} + {(\ln c)^3} - 3\ln a.\ln b.\ln c] = 0$
==> ${(\ln a)^3} + {(\ln b)^3} + {(\ln c)^3} - 3\ln a.\ln b.\ln c = 0$
==> $\ln a + \ln b + \ln c = 0$
==> $\ln (abc) = ln 1$, $[{a^3} + {b^3} + {c^3} - 3abc = 0$
==> $a + b + c = 0]$,
$\therefore abc = 1$.
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 $3.$
(where [.], {.} and $sgn\ x$ denotes greatest integer function, fractional part function and signum function respectively)