Maths geeks on Nairaland.

Please help me with this question

In an AP, the sum of three term of positive number is 21 while the sum of the square number is 155. Find the number.

Very simple

Kylekent59:
Very simple

Drop the solution plz

the three terms... are they consecutive?
don't think enough information was given

Assuming the numbers follow each other,

Your numbers are 5, 7 and 9.

5+7+9= 21

25+49+81= 155

Let x represent the middle terms
Let d represent the common difference
I.e (x-d)+ x + (x+d) = 21... the sum of the positive numbers... equ (1)
And also
(x-d)^2 + x^2 + (x+d)^2 = 155.... sum of their squares...equ (2)
From equ.. 1
Expanding equ..1
X-d+x+x+d= 21
3x=21
X= 7
Also fron the sum of their squares from equ (2)
Expanding u get
X^2-2xd+d^2+x^2+x^2+2xd+d^2 = 155
3x^2 + 2d^2 = 155... equ (3)
But we know x to be 7 from equ 1
So we put 7 for x in equ..(3)
3(7)^2 + 2d^2 =155
147 + 2d^2 = 155
2d^2 =155-147
d^2 = 4
d=2
So to get the three consecutive terms we substitutes the values for d and x into equ 1 and 2
From equ..(1)
(x-d) + x + (x+d) =21
First (x-d)= 7-2 = 5
Second is x = 7
The thirs is (x+d)= (7+2)= 9
So the three numbers are 5,7 and 9
To check if we are correct
(x-d) + x + (x+d) = 21
5+7+9= 21
Also for their squares
25+49+81=155
So the ans is correct
5,7, and 9.
Still me kylekent59.

Lalastica plz move this to front page. It is gona help a lot of people finding it difficult in this topic called arithmetic progression(A.P)