Sum $2 a - b, 4 a - 3 b, 6 a - 5 b$ , .... to $n$ terms.

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Solution

The given series are $(2 a - b), (4 a - 3 b), (6 a - 5 b), . . . n$ terms
$∴ a = (2 a - b)$
and $d = ((4 a - 3 b) - (2 a - b)) = 2 a - 2 b = 2 (a - b)$
$∴ S_{n} = 2 a + (n - 1) d$
$⟹ S_{n} = 2 (2 a - b) + [(n - 1) \times 2 (a - b)]$
${⟹ S}_{n} = (4 a - 2 b) + [(2 a - 2 b) \times (n - 1)]$
$= 4 a - 2 b - 2 a + 2 b + 2 a n - 2 b n$
$= 2 a + 2 a n - 2 b n$
$S_{n} = 2 [a (2 n + 1) - b n]$

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