Question

find x,so that 2x+1 ,x^2 +x+1 and 3x^2 -3x+3 are consecutive terms of an A.P.​

Answer 1

Answer:

$\sf{The \ values \ of \ x \ are \ 1 \ and \ 2.}$

Given:

• $\sf{In \ an \ AP,}$

• $\sf{2x+1, \ x²+x+1 \ and \ 3x²-3x+3}$ $\sf{are \ consecutive \ terms.}$

Solution:

$\sf{Here,}$

$\sf{t_{1}=2x+1,}$

$\sf{t_{2}=x^{2}+x+1,}$

$\sf{t_{3}=3x^{2}-3x+3}$

$\sf{we \ know,}$

$\boxed{\sf{2\times \ t_{2}=t_{1}+t_{3}}}$

$\sf{\therefore{2(x^{2}+x+1)=(2x+1)+(3x^{2}-3x+3}}$

$\sf{\therefore{2x^{2}+2x+2=3x^{2}-x+4}}$

$\sf{\therefore{x^{2}-3x+2=0}}$

$\sf{\therefore{x^{2}-x-2x+2=0}}$

$\sf{\therefore{x(x-1)-2(x-1)=0}}$

$\sf{\therefore{(x-1)(x-2)=0}}$

$\sf{\therefore{x=1 \ or \ 2}}$

$\sf\purple{\tt{\therefore{The \ values \ of \ x \ are \ 1 \ and \ 2.}}}$

_____________________________

$\sf\blue{Information:}$

$\sf{we \ know, \ t_{2}=a+d}$

$\sf{\therefore{2\times \ t_{2}=2a+2d...(1)}}$

$\sf{Also, \ t_{1}=a \ and \ t_{3}=a+2d}$

$\sf{\therefore{t_{1}+t_{3}=(a)+(a+2d)}}$

$\sf{\therefore{t_{1}+t_{3}=2a+2d...(2)}}$

$\sf{...from \ (1) \ and \ (2), \ we \ get}$

$\sf{\longmapsto{2\times \ t_{2}=t_{1}+t_{3}}}$