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8. Sequences and Series
easy
If $x$ is added to each of numbers $3, 9, 21$ so that the resulting numbers may be in $G.P.$, then the value of $x$ will be
A
$3$
B
$\frac{1}{2}$
C
$2$
D
$\frac{1}{3}$
Solution
(a) $3 + x,\;9 + x,\;21 + x$ are in $G.P.$
$ \Rightarrow $ ${(9 + x)^2} = (3 + x)(21 + x)$
$ \Rightarrow $ $81 + {x^2} + 18x = {x^2} + 24x + 63$
$ \Rightarrow $ $6x = 18$ or $x = 3$.
Trick : Check for $(a)$, $3 + 3,\;9 + 3,\;21 + 3$ are in $G.P.$
Standard 11
Mathematics
