Answer :

P(n): a + (a + d) + (a + 2d) + … + (a + (n– 1)d) =


For n = 1


a = [2a + (1 - 1)d]


a = a


Since, P(n) is true for n =1,


Let P(n) is true for n = k, so


a + (a + d) + (a + 2d) + … + (a + (k– 1)d) = - - - - - (1)


We have to show that,


a + (a + d) + (a + 2d) + … + (a + (k– 1)d) + (a + (k)d) = [2a + kd]


Now,


{a + (a + d) + (a + 2d) + … + (a + (k– 1)d)} + (a + kd)


= [2a + (k - 1)d] + (a + kd) using equation


=


=


=


=


= [2a + kd]


Therefore, P(n) is true for n = k + 1


Hence, P(n) is true all n N by PMI


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