Question

Determine 'a' so that 2a+1 , a^2+a+1 and 3a^2-3a+3 are consecutive terms.

Answer 1

d1=2A+1

d2=A2+A+1

d3=3A2-3A+3

 

as acoording to arithmatic progression common difference remain same

 

therefore

d2-d1=d3-d2

A2+A+1-(2A+1)=3A2-3A+3-(A2+A+1)

A2+A+1-2A-1=3A2-3A+3-A2-A-1

A2-A=2A2-4A+2

0=A2-3A+2

0=A2-2A-A+2

0=A(A-2)-1(A-2)

0=(A-1)(A-2)

A=1 or A=2

follow this

Answer 2

Answer:

Step-by-step explanation:

d1=2A+1

d2=A2+A+1

d3=3A2-3A+3

 

as acoording to arithmatic progression common difference remain same

 

therefore

d2-d1=d3-d2

A2+A+1-(2A+1)=3A2-3A+3-(A2+A+1)

A2+A+1-2A-1=3A2-3A+3-A2-A-1

A2-A=2A2-4A+2

0=A2-3A+2

0=A2-2A-A+2

0=A(A-2)-1(A-2)

0=(A-1)(A-2)

A=1 or A=2