Answer :

We have,

Second term, a_{2} = 7

⇒ a + d = 7 [1]

Fourth term, a_{4} = 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. , t_{n} = a + (n – 1) d.

First, t_{n} = a + (n – 1) d

Then, t_{m} = a + (m – 1) d

Given, mt_{m} = nt_{n}

⇒ m [a + (m – 1) d] = n [a + (n – 1) d]

⇒ ma + m^{2}d – md = na + n^{2}d – nd

⇒ ma – na + m^{2}d – n^{2}d – md + nd = 0

⇒ a (m – n) + d (m^{2} – n^{2}) – d (m – n) = 0

We know that a^{2} – b^{2} = (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, t_{m + n} = 0

Hence proved.

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