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Mathematics

The nth term of the A.P. 1m,(1+m)m,(1+2m)m\dfrac {1}{m}, \dfrac{(1 + m)}{m}, \dfrac{(1 + 2m)}{m}, … is:

  1. m(n−1)+1m\dfrac{m(n − 1) + 1}{m}

  2. m(n+1)−1m\dfrac{m(n + 1) − 1}{m}

  3. m(n−1)−1m\dfrac{m(n − 1) − 1}{m}

  4. None of these

AP

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Answer

Given,

a = 1m\dfrac{1}{m}

⇒d=(1+m)m−1m⇒d=(1+m)−1m⇒d=mm⇒d=1.\Rightarrow d = \dfrac{(1 + m)}{m} - \dfrac {1}{m} \\[1em] \Rightarrow d = \dfrac{(1 + m) - 1}{m} \\[1em] \Rightarrow d = \dfrac{m}{m} \\[1em] \Rightarrow d = 1.

We know that,

nth term of A.P. :

∴ Tn = a + (n - 1)d

⇒Tn=1m+(n−1)×1=1+m(n−1)m=m(n−1)+1m.\Rightarrow T_n = \dfrac{1}{m} + (n - 1) \times 1 \\[1em] = \dfrac{1 + m(n - 1)}{m} \\[1em] = \dfrac{m(n - 1) + 1}{m}.

Hence, option 1 is the correct option.

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