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Q.
Consider the ten numbers $ar, ar^{2}, ar^{3}, ..., ar^{10}.$ If their sum
is 18 and the sum of their reciprocals is 6, then the product of these ten numbers is
Sequences and Series
Solution:
Given $\frac{ar(r^{10}-1)}{r-1}=18\,...(1)$
Also $\frac{\frac{1}{ar}(1-\frac{1}{r^{10}})}{1-\frac{1}{r}}=6$
or $\frac{1}{ar^{11}} \cdot \frac{(r^{10}-1)r}{r-1}=6$
or $\frac{1}{a^{2}r^{11}} \cdot\frac{ar(r^{10}-1)}{r-1}=6\,...(2)$
From (1) and (2),
$\frac{1}{a^{2}r^{11}} \times 18=6$
or $a^{2}r^{11} =3$
Now $P=a^{10}r^{55} =(a^{2}r^{11})^{5} $
$=3^{5} =243$