Question

If the nth term of a progression be a linear expression in n then
prove that this progression is an AP.

Answer 1

Answer:

Let the progression be tn

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According to the question: tn=an+b

Let us take the consecutive difference: tn−tn−1 =an+b−a(n−1)−b=2a

As the consecutive difference is constant, the sequence is an AP by definition of an AP.

Step-by-step explanation:

Answer 2

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$let \: the \: nth \: term \: of \: a \: given \: progression \: be \: given \: by \\ t _{n} = an + b \: \: where \: a \: and \: \: b \: \: are \: constant \\ then \: t _{n - 1} = a(n - 1) + b = (an + b) - a \\ \therefore \: t _{n} - t _{n - 1} = (an + b) - (an + b) + a = a \: \\ which \: is \: constant \\ hence \: the \: given \: progression \: is \: an \: ap$