Determine 'a' so that 2a+1 , a^2+a+1 and 3a^2-3a+3 are consecutive terms.
Answers
Answered by
49
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
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
Answered by
15
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
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