Question

Find four consecutive terms in A.P whose sum is 20 and the sum of whose squares is 120.

Answer 1

Let the ap be a-3d, a-d, a+d, a+3d
so according to the question
a-3d+a-d + a+d + a+3d =20
4a =20
a=5
also
$(a-3d{)}^2+(a-d{)}^2+(a+d{)}^2+(a+3d{)}^2=120$
$(5-3d{)}^2+(5-d{)}^2+(5+d{)}^2+(5+3d)=120$
$25+9d^2-30d+25+d^2-10d+25+d^2+10d+25+9d^2+30d=120$
$100+20d^2=120$
$20d^2=120-100$
$d^2=1$
$d=1$
therefore the AP will be 5-3(1), 5-1, 5+1, 5+3(1)
2,4,6,8