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8. Sequences and Series
easy
If the ${p^{th}}$,${q^{th}}$ and ${r^{th}}$ term of a $G.P.$ are $a,\;b,\;c$ respectively, then ${a^{q - r}}{b^{r - p}}{c^{p - q}}$ is equal to
A
$0$
B
$1$
C
$abc$
D
$pqr$
Solution
(b) Let $AR^{p-1}=a$ …..(i)
$A{R^{q – 1}} = b$ …..(ii)
and $A{R^{r – 1}} = c$ …..(iii)
So ${a^{q – r}}{b^{r – p}}{c^{p – q}}$ $ = {\left\{ {A{R^{p – 1}}} \right\}^{q – r}}\left\{ {A{R^{q – 1}}} \right\}{\,^{r – p}}{\left\{ {A{R^{r – 1}}} \right\}^{p – q}}$
$ = {A^0}{R^0} = 1$.
Note : Such type of questions $i.e.$ containing terms of powers in cyclic order associated with negative sign, reduce to $1 $ mostly.
Standard 11
Mathematics
