Question

Find five numbers in Arithmetic progression whose sum is 25 and the sum of whose squares as 135.

Answer 1

Let the numbers lee

a-2d,a-d,a,a+d,a+2d.

$\Rightarrow a-2d,+a-d+a+a+a+a+2d=25$

$\Rightarrow 5a=25\Rightarrow a=5$

$\left[\right(a-2d{)}^2+(a-d{)}^2+a^2(a+d{)}^2+(a+2d{)}^2]=135$

$\Rightarrow (5-2d{)}^2+(5d{)}^2+(5+d{)}^2(5+2d{)}^2=135$

$\left[\right(a+b{)}^2+(a-b{)}^2=2(a^2+b^2\left)\right]$

$\Rightarrow 2(25+4d{)}^2+2(25+d{)}^2=110$

$\Rightarrow 25+4d^2+25+d^2=55$

$\Rightarrow 5d^2=5$

$\Rightarrow d^2=1$ $d=\pm 1$

Numbers =3,4,5,6,7 or 7,6,5,4,3