Question

The sum of three consecutive terms of gp is 7 and the sum of their squares is 21 find the numbers

Answer 1

Step-by-step explanation:

x+(x+1)+(x+2)=7

3x+3=7

3x=7-3

3x=4

x=4/3

1st no=x=4/3

2nd no=x+1=4/3+1=7/3

3rd no=x+2=4/3+2=10/3

1st+2nd+3rd=4/3+7/3+10/3=21/3=7

2)

x^2+(x+1)^2+(x+2)^2=21

x^2+x^2+1+2x+x^2+4+4x=21

3x^2+6x+5=21

3x^2+6x=21-5

3x^2+6x=16

3x(x+2)=16

3x=16,x+2=16

x=16/3,x=16-2

x=16/3,x=14

1st nob=x=16/3,1st=x=14

2nd nob=(16/3+1)^2,2st=(14+1)^2

=361/9,=225

3rd=(16/3+2)^2,,2nd=(14+2)^2

=484/8,,=256