Question

The sum of four numbers which form an arithmetic progression is $1$ and the sum of the squares of those numbers is $0.3$. Find the numbers.

Answer 1

Let the four numbers be $(a-3d),(a-d),(a+d)$ and $(a+3d)$

$\therefore a-3d+a-d+a+d+a+3d=1\Rightarrow 4a=1\Rightarrow a=\frac{1}{4}-\left(i\right)$

Sum of squares of $4$ numbers$=0.3$

$\Rightarrow {(a-3d)}^2+{(a-d)}^2+{(a+d)}^2+{(a+3d)}^2=0.3\Rightarrow a^2-6d+9d^2+a^2-2d+d^2+a^2+2d+d^2+a^2+6d+{9d}^2=0.3\Rightarrow 4a^2+20d^2=0.3\Rightarrow a^2+5d^2=\frac{0.3}{4}=0.075$

Replacing $a=\frac{1}{4}=0.25$

$\Rightarrow {\left(0.25\right)}^2+5d^2=0.075\Rightarrow {5d}^2=0.075-{\left(0.25\right)}^2=0.075-0.0625=0.0125\Rightarrow d^2=\frac{0.0125}{5}=\frac{125}{5\times 10000}=\frac{25}{10000}\Rightarrow d=\frac{5}{100}=0.05$

$\therefore$The numbers are $a-3d,a-d,a+d,a+3d$

$=0.25-3\times \mathrm{0.050.25}-0.05,0.25+0.05,0.25+3\times 0.05=0.10,0.20,0.30,0.40$