Question

Five numbers are in continued proportion. The product of the first and the fifth is $256$ and the sum of the second and the fourth is $40$. Find the numbers.

Answer 1

Let the number be $a,b,c,d,e$ such that $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}=k$

$\Rightarrow d=ek,c=dk=e{k}^2,b=ck=c{k}^3,a=bk=e{k}^4$ ___ (1)

According to conditions, $a\times e=256$

$\Rightarrow e{k}^4\times e=256$

$e{k}^2=16$ ___ (2)

Also, $b+d=40\Rightarrow e{k}^3+ek=40$

$16k+\frac{16}{k}=40$ $[\because e{k}^2=16]$

$16k^2-40k+16=0$

$4k^2-10k+4=0$

$k=\frac{10\pm \sqrt{100-43}}{8}=\frac{10\pm 6}{8}=2,\frac{1}{2}$

When $k=2,e=\frac{16}{4}=4,d=4\left(2\right)=8,c=4\left(4\right)=16$

$b=4\left(8\right)=32,a=4\left(16\right)=64$

When $k=\frac{1}{2},e=16\times 4=64,d=32,c=16,b=8,a=4$

$\therefore$ Numbers are $4,8,16,32,64$ or $64,32,16,8,4$