Answer to Question #122821 in Math for Ojugbele Daniel

Question

The sum of five numbers in arithmetic progression is 25 and of their square is 165. Find the numbers.

Answer 1

$a_1+a_1+d+a_1+2d+a_1+3d+a_1+4d=25$

$a_1^2+(a_1+d)^2+(a_1+2d)^2+$ $+(a_1+3d)^2+(a_1+4d)^2=165$

$a_1=5-2d$

$(5-2d)^2+(5-d)^2+(5)^2+(5+d)^2+(5+2d)^2=165$

$25-20d+4d^2+25-10d+d^2+25+$ $+25+10d+d^2+25+20d+4d^2=165$

$10d^2=40$

$d^2=4$

$d_1=-2, d_2=2$

$d=-2$

$a_1=5-2(-2)=9$

The numbers are $9,7,5,3,1.$

$d=2$

$a_1=5-2(2)=1$

The numbers are $1,3,5,7,9.$