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Determine a so that 2a+1 , a­2+a+1 and 3a2-3a+3 are consecutive terms in an A.P

Dear Student!

 

@sentia : Your friend has provided correct answer! Hope you got it!

@setu06 : Good Work! Keep it up & Keep Posting!

 

Cheers!

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For 2a+1, a2+a+1, and 3a2-3a+3 to be the consecutive terms in an A.P. their common difference should be the same.

So, (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

  a2 - 3a + 2 = 0

 (a - 2) (a - 1) = 0

 a = 2 or a = 1

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setu thanks a lot

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