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8. Sequences and Series
easy
If the sum of first 6 term is $9$ times to the sum of first $3$ terms of the same $G.P.$, then the common ratio of the series will be
A
$ - 2$
B
$2$
C
$1$
D
$1/2$
Solution
(b) Under given conditions, we get
$ \Rightarrow $ $\frac{{a({r^6} – 1)}}{{(r – 1)}} = 9.\frac{{a({r^3} – 1)}}{{(r – 1)}}$ $(\because \;r > 1)$
$ \Rightarrow $ ${r^6} – 1 = 9{r^3} – 9$
$ \Rightarrow $ ${({r^3})^2} – 9({r^3}) + 8 = 0$
$ \Rightarrow $ $({r^3} – 1)({r^3} – 8) = 0$
$ \Rightarrow $ $r = 1,\;\omega ,\;{\omega ^2}$
and $r = 2$. But $r = 1,\;\omega ,\;{\omega ^2}$ cannot satisfy the given condition.
Hence $r = 2$.
Standard 11
Mathematics
