Question

class 11 maths A.P SERIES​

Answer 1

$\large \bold \green{answer \: = \: 7q - 6p }$

$\large \bold{ \underline{given}} \\ third \: \: term \: \: of \: \: ap \: \: is = p \\ fourth \: \: term \: \: of \: \: ap \: is = q \\ \\ \large \bold { \underline{to \: find}} \\ 10 \: \: terms \: \: and \: \: general \: \: terms \\ \large \bold \green{ \underline{explanation}} \\ T3 = a +2d = p.....(1) \\ T4 = a + 3d = q....(2) \\ from \: \: equation \: \: (1) \: \: and \: \: (2) \: \: we \: \: get \\ d = q - p \\ put \: \: value \: \: of \: \: d \: \: in \: \: equation \: \: (1) \\ a + 2(q - p) = p \\ a + 2q - 2p = p \\ a = 3p - 2q \\ \large \bold \blue{ \underline{general \: \: term}} \\ an = a + (n - 1)d \\ an = (3p - 2q) + (n - 1)(q - p) \\ an = 3p - 2q - q + p + n(q - p) \\ an = 4p - 3q + n(q - p) \\ \large \bold \orange{ \underline{10 \: \: terms \: \: of \: \: ap}} \\ put \: n = 10 \\ 4p - 3q + 10(q - p) \\ 4p - 3q + 10q - 10p \\ 7q - 6p$