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If the lines $x = k; k = 1, 2 \dots, n$ meet the line $y = 3 x + 4$ at the points $A_{k} (x_{k}, y_{k}), k = 1, 2, \dots, n$ then
the ordinate of the centre of Mean position of the points $A_{k}, k = 1, 2, \dots, n$ is

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n+12

3n+112

3(n+1)2

none of these

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detailed solution

Correct option is B

We have yk=3k+4 the ordinate of Ak, the point of intersection of x=k and y=3x+4. So the ordinate of the centre of Mean position of the points Ak,k=1,2,…n is1n∑k=1n yk=1n∑k=1n (3k+4)=3n∑k=1n k+4=3n(n+1)n⋅2+4=3n+112.

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If the lines x=k;k=1,2…,n meet the line y=3x+4 at the points Akxk,yk,k=1,2,…,n thenthe ordinate of the centre of Mean position of the points Ak,k=1,2,…,n is

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If the lines $x = k; k = 1, 2 \dots, n$ meet the line $y = 3 x + 4$ at the points $A_{k} (x_{k}, y_{k}), k = 1, 2, \dots, n$ then
the ordinate of the centre of Mean position of the points $A_{k}, k = 1, 2, \dots, n$ is