find numbers in A.P such that their sum is 24 and the sum of their squares is 200
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Sn=n/2[2a +(n-1)d]
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Let us assume that the numbers in AP are a, a+d and a-d ( a is a specific number and d is difference between them)
given a + (a+d) + (a-d) = 24
3 a=24
a=8
Also a² + (a+d)² + (a-d)² = 200
By solving we get 3a² + 2d² = 200
placing a=8
2d² = 200 - 192
d = 2
Therefore substituting a= 8 and d = 2 we get 6, 8,10
given a + (a+d) + (a-d) = 24
3 a=24
a=8
Also a² + (a+d)² + (a-d)² = 200
By solving we get 3a² + 2d² = 200
placing a=8
2d² = 200 - 192
d = 2
Therefore substituting a= 8 and d = 2 we get 6, 8,10
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