Determine 'a' so that $2 a + 1, a^{2} + a + 1$ and ${3 a}^{2} - 3 a + 3$ are consecutive terms of an A .P.

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Solution

If a, b, c are in AP then $2 b = a + c$
So,
$2 (a^{2} + a + 1) = 3 a^{2} - 3 a + 3 + 2 a + 1$
$2 a^{2} + 2 a + 2 = 3 a^{2} - a + 4$
$\Rightarrow 3 a^{2} - 2 a^{2} - a - 2 + 4 - 2 = 0$
$\Rightarrow a^{2} - 3 a + 2 = 0$
$\Rightarrow a^{2} - 2 a - a + 2 = 0$
$\Rightarrow a (a - 2) - 1 (a - 2) = 0$
So $(a - 1) (a - 2) = 0$
So $a = 1$ or $a = 2$ .

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