Question

If the numbers (x-2), (4x-1). (5x+2) are in A.P. find x.​

Answer 1

EXPLANATION.

Numbers = ( x - 2), ( 4x - 1 ), ( 5x + 2 ) .....Ap.

Method = 1.

Common difference of an Ap = b - a = c - b.

→ ( 4x - 1 ) - ( x - 2) = ( 5x + 2) - ( 4x - 1).

→ 4x - 1 - x + 2 = 5x + 2 - 4x + 1.

→ 3x + 1 = x + 3.

→ 3x - x = 3 - 1.

→ 2x = 2.

→ x = 1.

Method = 2.

Conditions of an Ap → 2b = a + c.

→ 2( 4x - 1 ) = ( x - 2 ) + ( 5x + 2).

→ 8x - 2 = x - 2 + 5x + 2.

→ 8x - 2 = 6x.

→ 8x - 6x = 2.

→ 2x = 2.

→ x = 1.

Value of x = 1.

Answer 2

$\huge \sf \underline \red{Answer : }$

$\sf \therefore{ \boxed{ \underline {\underline{ \purple{ \sf{value \: of \: x \: is \: 1}}}}}}$

$\implies \sf \huge \underline \blue{Given : }$

• if the number (x-2),(4x-1),(5x+2) are in Ap

$\implies \sf \huge \underline \pink{To \: find : }$

• x = ?

$\implies \huge \sf \underline \orange{solution : }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ (x - 2) \: (4x - 1) \: (5x + 2) are \: in \: Ap}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \star \sf{(4x - 1) - (x - 2) = (5x + 2) - (4x - 1)}$

$\sf \underline{this \: is \: difference \: between \: two \: consectuvie \: term \: so \: this \: is \: same}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \star \sf{4x - 1 - x + 2 = 5x + 2 - 4x + 1}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \star \sf{3x + 1 = x + 3}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \star \sf{3x - x = 3 - 1 }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \star \sf{2x = 2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \star \sf{x = 1}$

$\tt \underline \red{x = 1}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{or}$

$\: \: \sf \implies{(x - 2 )\:( 4x - 1 )\: and \: (5x + 2 )\: are \: in \: ap}$

$\: \: \: \: \: \: \: \: \: \: \: \sf \implies{2(4x - 1) = x - 2 + 5x + 2}$

$\: \: \: \: \: \: \: \: \: \: \: \sf \implies{8x - 6x = 2}$

$\: \: \: \: \: \: \: \: \: \: \: \sf \implies{2x = 2}$

$\: \: \: \: \: \: \: \: \: \: \: \sf \implies{x = 1}$

$\sf \therefore \underline{value \: of \: x \: is \: 1}$