Question

Find the value ofx for which (x + 2), 2x, (2x + 3) are three consecutive terms of an AP.​

Answer 1

EXPLANATION.

To find the value of x.

Condition of an Ap = 2b = a + c.

→ ( x + 2 ),2x , ( 2x + 3 )

→ 2(2x) = ( x + 2 ) + ( 2x + 3 ).

→ 4x = 3x + 5.

→ 4x - 3x = 5.

→ x = 5.

Method = 2.

in Ap Common difference = b - a = c - b.

→ 2x - ( x + 2 ) = ( 2x + 3 ) - ( 2x ).

→ 2x - x - 2 = 2x + 3 - 2x.

→ x - 2 = 3.

→ x = 5.

More information.

First term of an Ap = a

Common difference = d

Nth term of an Ap = An = a + ( n - 1 )d.

Answer 2

Given,

$\sf (x+2),\;2x,\;(2x+3)\;are\;three\; consecutive\;terms\;of\;an\;AP$

Now, there are two possible ways to find the value of x.

$\sf First\;way$

$\sf We\;know\;that\;common\;difference\;in\;AP\;is\;same.$

$\sf So,$

$\sf \to(2x)-(x+2)=(2x+3)-(2x)$

$\sf \to2x-x-2=2x+3-2x$

$\sf \to x-2=3$

$\sf \to x=5$

$\sf Second\;way$

$\sf If\;a,\;b\;and\;c\;are\;three\;consecutive\;terms\;of\;an\;AP,$

$\sf 2b=a+c$

$\sf Here,$

$\sf \to a=(x+2)$

$\sf \to b=(2x)$

$\sf \to c=(2x+3)$

$\sf \to2(2x)=(x+2)+(2x+3)$

$\sf \to4x=3x+5$

$\sf \to4x-3x=5$

$\sf \to x=5$