Find 4 numbers which are in AP whose sum is 20 and sum of their squires is 120
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S(4)=20
a+(a+d)+(a+2d)+(a+3d)=20
4a+6d=20
diving by 2 on both sides of the equation,
2a+3d=10
2a=10-3d
a=(10-3d)/2 - - - - - - - - (1)
a²+(a+d)²+(a+2d)²+(a+3d)²=120
a²+(a²+d²+2ad)+(a²+4d²+4ad)+(a²+9d²+6ad)=120
4a²+14d²+12ad=120
diving by 2 on both sides of the equation,
2a²+7d²+6ad=60 - - - - - - - - (2)
Substituting (1) in (2) and solving we will get value of d
Then substitute value of d in equation (1) to get the value of a
Then put values of a&d in
a, (a+d), (a+2d), (a+3d) to get the four numbers
a+(a+d)+(a+2d)+(a+3d)=20
4a+6d=20
diving by 2 on both sides of the equation,
2a+3d=10
2a=10-3d
a=(10-3d)/2 - - - - - - - - (1)
a²+(a+d)²+(a+2d)²+(a+3d)²=120
a²+(a²+d²+2ad)+(a²+4d²+4ad)+(a²+9d²+6ad)=120
4a²+14d²+12ad=120
diving by 2 on both sides of the equation,
2a²+7d²+6ad=60 - - - - - - - - (2)
Substituting (1) in (2) and solving we will get value of d
Then substitute value of d in equation (1) to get the value of a
Then put values of a&d in
a, (a+d), (a+2d), (a+3d) to get the four numbers
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