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Q.
If the sum of three numbers in G.P. is $63 $ and the product of the first and the second term is $\frac{3}{4}$ of the third term, then the numbers are
Sequences and Series
Solution:
Let the three numbers be $a, a r, a r^{2}$
Given $a+a r+a r^{2}=63$
and $a \cdot a r=\frac{3}{4} \cdot a r^{2}$
or $a=\frac{3}{4} r$
Putting in (1), $\frac{3}{4} r+\frac{3}{4} r \cdot r+\frac{3}{4} r \cdot r^{2}=63$
or $r^{3}+r^{2}+r-84=0$
or$\therefore r=4, \frac{-5 \pm \sqrt{25-84}}{2}$
$\therefore $ Real value of $r$ is $4$.
Putting this value in (2),
$a=\frac{3}{4} \times 4=3$
$\therefore $ The three numbers are, $3,3 \times 4,3 \times 4^{2}$,
i.e., $3,12,48$.