Question

the sum of all numbers is 204 the average of first two numbers is 56 the third number is 54 find the force number

Answer 1

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${BY AP :}$

$let the no. are$

$(a-3d),(a-d),(a+d), &(a+3d)$

$SUM OF ALL NO. IS 204$

=>4a=204

=>a=51

°•°AVERAGE OF FIRST TWO NO. :

=>[(a-3d)+(a-d)]/2=56

=>2a-4d=56

=>a-2d=56••••••[1]

°•°3rd NO. IS 56

=>(a+d)=56••••[2]

◆From eq [1] and [2]

we have:

a-2d=56
a+d=56
-...-...-
______
-3d=0

=>d=0

•°• FORTH NO. IS (a+3d) =>(51+3×0)

=>51

★$check :$

a-3d =51

a+3d=51

a-d=51

a+d=51

total=204

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