the end term of an ap an=24 -3 and then the second term is
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We know that the nth term of an AP is an = a + (n – 1)d Where, a = first term an is nth term d is the common difference a2 = a + d = 13 …..(1) a5 = a + 4d = 25 …… (2) From equation (1) we have, a = 13 – d Using this in equation (2), we have 13 – d + 4d = 25 13 + 3d = 25 3d = 12 d = 4 a = 13 – 4 = 9 a7 = a + 6d = 9 + 6(4) = 9 + 24 = 33
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