Answer :

We have,

Second term, a2 = 7


a + d = 7 [1]


Fourth term, a4 = 23


a + 3d = 23 [2]


[2] – [1] gives,


3d – d = 23 – 7


2d = 16


d = 8


a + 8 = 7


a = -1


b = 7 + d = 7 + 8 = 15


c = 23 + d = 23 + 8 = 31


OR


We know that nth term of A.P. , tn = a + (n – 1) d.


First, tn = a + (n – 1) d


Then, tm = a + (m – 1) d


Given, mtm = ntn


m [a + (m – 1) d] = n [a + (n – 1) d]


ma + m2d – md = na + n2d – nd


ma na + m2d – n2d – md + nd = 0


a (m n) + d (m2 – n2) – d (m – n) = 0


We know that a2 – b2 = (a – b) (a + b)


(m n) [a + d (m + n) d] = 0


[a + d (m + n) d] = 0


a + (m + n 1) d = 0


(m + n)th term, tm + n = 0


Hence proved.


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