math solution of class 11 exercise 15.2 of kc sinha book
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let the numbers be a-d,a,a+d
then
sum of numbers
=a-d+a+a+d=27
3a= 27
a=27/3 =9
so a=9
now sum of there squares
(a-d)^2+a^2+(a+d)^2=275
i.e
[(3-d)2]+9+[(3+d ^2)]=275
(9+d^2-6d)]+9+(9+d^2+6d)=275
2d^2+27=275
2d^2=275-27
2d^2=248
d^2=248/2= 124
d=√124
d=2√31
therefore the no are
(3-2√31); (3); (3+2√31)
then
sum of numbers
=a-d+a+a+d=27
3a= 27
a=27/3 =9
so a=9
now sum of there squares
(a-d)^2+a^2+(a+d)^2=275
i.e
[(3-d)2]+9+[(3+d ^2)]=275
(9+d^2-6d)]+9+(9+d^2+6d)=275
2d^2+27=275
2d^2=275-27
2d^2=248
d^2=248/2= 124
d=√124
d=2√31
therefore the no are
(3-2√31); (3); (3+2√31)
Avinash2000:
is it right
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