find four consecutive terms in A.P. whose sum is 20 and the sum of square is 120.
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Let the four numbers in A.P be a-3d, a-d,a+d,a+3d. ---- (1)
Given that Sum of the terms = 20.
= (a-3d) + (a-d) + (a+d) + (a+3d) = 20
4a = 20
a = 5. ---- (2)
Given that sum of squares of the term = 120.
= (a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 120
= (a^2 + 9d^2 - 6ad) + (a^2+d^2-2ab) + (a^2+d^2+2ad) + (a^2+9d^2+6ad) = 120
= 4a^2 + 20d^2 = 120
Substitute a = 5 from (2) .
4(5)^2 + 20d^2 = 120
100 + 20d^2 = 120
20d^2 = 20
d = +1 (or) - 1.
Since AP cannot be negative.
Substitute a = 5 and d = 1 in (1), we get
a - 3d, a-d, a+d, a+3d = 2,4,6,8.
let four consecutive terms be (a-3d),(a-d),(a+d) and (a+3d)
given,
a-3d+a-d+a+d+a+3d=20
4a=20
a=5
also
(5-3d)²+(5-d)²+(5+d)²+(5+3d)²=120
25+9d²-30d+25+d²-10d+25+d²+10d+25+9d²+30d=120
100+20d²=120
20d²=20
d²=1
d=±√1
d=±1
thus four consecutive terms are
2,4,6,8
and
8,6,4,2