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third term of an ap is p and fourth term is q then the 10 tern is
Dear studentA3 = p = a + 2da is first term d is common differenced = (q-p)A4 = a + 3d = qa= 3p-2qA10 = a + 9d(3p-2q) +(q-p)9= 7q -6p
Given t3 = p and t4 = qnth term or general term of AP is tn = a + (n – 1)d⇒ t3 = a + 2d = p → (1)⇒ t4 = a + 3d = q → (2)Subtract (1) from (2), we getd = q – pPut value of d in (1), we geta + 2(q – p) = p⇒ a = p – 2q + 2p = 3p – 2qGeneral term, tn = (3p – 2q) + (n – 1)(q – p)= 3p – 2q – q + p + n(q – p)tn = 4p – 3q + n(q – p)Put n = 10 in tnt10 = 4p – 3q + 10(q – p) = 4p – 3q + 10q – 10p = 7q – 6p
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