Question

in an a.p the 10th term is 31 and 15th term is 46. find the series

Answer 1

$a10 = 31 \\ a + 9d = 31 \\ \\ a15 = 46 \\ a + 14d = 46 \\ \\ a + 14d = 46 \\ a + 9d = 31 \\ subtract \: \: the \: equations \\ 5d = 15 \: \\ d = 3 \\ as \: d = 3 \: substitute \: it \: in \: equation \: to \: find \: a \\ a + 9d = 31 \\ a = 31 - 9d \\ a = 31 - 27 \\ a = 4 \\ ap \: is \: 4 \: 7 \: 10 \: 13$