The sum of three consecutive numbers of a G.P is 56. If we subtract 1, 7 and 21 from these numbers in order, the resulting numbers form an A.P. Find the numbers.
Let the three consecutive numbers in G.P be $\dfrac{a}{r}$, a, $ar$
where first term $= a$; common ratio $= r$
$\therefore$ $\dfrac{a}{r}+a+ar = 56$ $\implies$ $\dfrac{a}{r} + ar = 56 - a$ $\cdots$ (1)
On subtracting 1, 7 and 21 from these numbers in order gives $\dfrac{a}{r}-1$, $a-7$ and $ar-21$
Then, $\dfrac{a}{r}-1$, $a-7$, $ar-21$ are in A.P
Common difference $= a-7-\left(\dfrac{a}{r} - 1\right) = ar - 21 - \left(a-7\right)$
i.e. $a-\dfrac{a}{r}-6=ar-a-14$
i.e. $\dfrac{a}{r}+ar-2a= 8$
i.e. $\dfrac{a}{r} + ar = 8 + 2a$ $\cdots$ (2)
$\therefore$ From equations (1) and (2) we have
$56-a = 8 + 2a$ i.e. $3a = 48$ $\implies$ $a=16$
Therefore, from equation (2) we have,
$\dfrac{16}{r} + 16r = 8 + 32 = 40$
i.e. $16r^2 -40r + 16 = 0$
i.e. $2r^2 -5r+2 = 0$
i.e. $\left(2r-1\right) \left(r-2\right) = 0$
i.e. $r = 2$ or $r = \dfrac{1}{2}$
Therefore, when $a=16$, $r=2$, the numbers are $\dfrac{16}{2}$, 16, $16\times2$ i.e. 8, 16, 32
When $a=16$, $r = \dfrac{1}{2}$, the numbers are $\dfrac{16}{1/2}$, 16, $16 \times \dfrac{1}{2}$ i.e. 32, 16, 8