Question

write the valve of x for which 2x , x+10 and 3x+2 are in ap​

Answer 1

$\large\sf\underline{Given:}$

Three terms :

• 2x

• x + 10

• 3x + 2

‎

$\large\sf\underline{To\:find:}$

• Value of x for which these terms are in AP .

‎

$\large\sf\underline{Solution:}$

We know ,

$\dag\:\underline{\sf In\:an\:AP\:,\:the\: common\: difference\:will\:be \:same.}$

So following this we get :

$\sf\implies\:x+10-2x=3x+2-(x+10)$

$\sf\implies\:x+10-2x=3x+2-x-10$

$\sf\implies\:x-2x+10=3x-x+2-10$

$\sf\implies\:-x+10=2x-8$

$\sf\implies\:-x-2x=-8-10$

$\sf\implies\:-3x=-18$

$\sf\implies\:\cancel{-}3x=\cancel{-}18$

$\sf\implies\:3x=18$

$\sf\implies\:x=\frac{18}{3}$

$\sf\implies\:x=\cancel{\frac{18}{3}}=6$

${\sf{{\pink{\implies\:x=6}}}}$

‎

So when x will be 6 the three terms 2x , (x + 10) and (3x + 2) will be at AP .

‎

Scroll left to right to view your answer properly !

!! Hope it helps !!