Question

# Find the value of x for which the numbers (5x + 2), (4x − 1) and (x + 2) are in AP.

Solution

## It is given that (5x + 2), (4x − 1) and (x + 2) are in AP. $\therefore \left(4x-1\right)-\left(5x+2\right)=\left(x+2\right)-\left(4x-1\right)\phantom{\rule{0ex}{0ex}}⇒4x-1-5x-2=x+2-4x+1\phantom{\rule{0ex}{0ex}}⇒-x-3=-3x+3\phantom{\rule{0ex}{0ex}}⇒3x-x=3+3$ $⇒2x=6\phantom{\rule{0ex}{0ex}}⇒x=3$ Hence, the value of x is 3.MathematicsRS Aggarwal (2018)Standard X

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