Question

what is the 12 th term of AP whose 4 th term is 65 and 8 th term is 129​

Answer 1

Answer:

$t4 = 65 \\ t8 = 129 \\ tn = a + (n - 1)d \\ for \: n = 4 \\ t4 = a + 3d \\ 65 = a +3d \\ for \: n = 8 \\ t8 = a + 7d \\ 129 = a + 7d \\ subtracting \: equation \: (1) \: from \: (2) \\ 4d = 64 \\ d = 16 \\ substituting \: the \: value \: of \: d \: in \: equation \: (1) \\ 65 = a + 3d \\ 65 = a + 3(16) \\ 65 = a + 48 \\ 65 - 48 = a \\ \\ a = 17$

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