Question

If the 7th term of an ap is 1 9 and its 9th term is 1 7 , find its 63rd term

Answer 1

i hope this will help u

Answer 2

$\huge \bf Solution :$

Let a be the first term of and d be the common difference of the given AP. Then,

$\bf T_{7} = 19$

$\bf \implies a + 6d = 19 - ①$

$\bf T_{9} = 17$

$\bf \implies a + 8d = 17 - ②$

$\bf On \: subtracting \: ① \: from \: ②, \: we \: get$

$\bf (a + 8d) - (a + 6d) = 17 - 19$

$\bf a + 8d - a - 6d = - 2$

$\bf 2d = - 2$

$\bf 2d = \dfrac{- 2}{2}$

$\bf d = \dfrac{- \cancel{2}}{\cancel{2}}$

$\bf d = - 1$

$\large \boxed{\bf d = - 1}$

$\bf By, \: putting \: d = - 1 \: in \: ①,$

$\bf we \: get$

$\bf a + 6 \times (- 1) = 19$

$\bf a - 6 = 19$

$\bf a = 19 + 6$

$\bf a = 25$

$\large \boxed{\bf a = 25}$

$\bf Thus, \: a = 25 \: and \: d = - 1$

$\bf \therefore T_{63} = a + (63 - 1)d = a + 62d$

$\bf \implies T_{63} = 25 + 62 \times (- 1)$

$\bf \implies T_{63} = 25 - 62$

$\bf \implies T_{63} = - 37$

$\bf Hence, \: 63rd \: term \: of \: given \: AP \: is \: -37.$