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8. Sequences and Series
medium
If $A = 1 + {r^z} + {r^{2z}} + {r^{3z}} + .......\infty $, then the value of $r$ will be
A
$A{(1 - A)^z}$
B
${\left( {\frac{{A - 1}}{A}} \right)^{1/z}}$
C
${\left( {\frac{1}{A} - 1} \right)^{1/z}}$
D
$A{(1 - A)^{1/z}}$
Solution
(b) $A = 1 + {r^z} + {r^{2z}} + {r^{3z}} + ……..\infty $
$A = 1 + [{r^z} + {r^{2z}} + {r^{3z}} + ……..\infty ]$
We know that sum of infinite $G.P.$ is
${S_\infty } = \frac{a}{{1 – r}}( – 1 < r < 1)$
Therefore, $A = 1 + \left[ {\frac{{{r^z}}}{{1 – {r^z}}}} \right]$
$\Rightarrow A = \frac{{1 – {r^z} + {r^z}}}{{1 – {r^z}}}$
$\therefore $ $A = \frac{1}{{1 – {r^z}}}$
$\Rightarrow 1 – {r^z} = \frac{1}{A} $
$\Rightarrow {r^z} = \frac{{A – 1}}{A}$
Hence $r = {\left[ {\frac{{A – 1}}{A}} \right]^{1/z}}$.
Standard 11
Mathematics
