Arithmetic progression

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Question

${If the fourth term of an arithmetic progression is 6, the m}^{t h} term is 18 and if the A.P. has integral terms only, then the number of such A.P.s is$

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Solution

$G i v e n t h a t T_{4} = a + 3 d = 6 a n d T_{m} = a + (m - 1) d = 18 (m - 4) d = 12 S i n c e t e r m s a r e i n t e g e r s, w e h a v e m - 4 = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 b u t m > 0 m - 4 = \pm 1, \pm 2, \pm 3, 4, 6, 12 . S o, m h a s n i n e p o s s i b l e v a l u e s$

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