Question

if t7 is equal to 1 by 9 and t9 is 1 by 7 find the 63rd term in AP

Answer 1

$t_7 = \frac{1}{9} \text{ and } t_9 = \frac{1}{7}$
We can write $t_7 = t_1 + 6d$ and $t_9 = t_1 + 8d$
$\therefore t_1 + 6d = \frac{1}{9} \text{ and } t_1 + 8d = \frac{1}{7}$
Solving this two equations we get :-
$t_1 = \frac{1}{63} \text{ and } d = \frac{1}{63}$

Thus if we need to find the 63$^{rd}$ term in AP...

$t_{63} = t_1 + 62d$
$t_{63} = \frac{1}{63} + 62 \cdot \frac{1}{63}$
$t_{63} = \frac{1}{63} (1 + 62)$
$t_{63} = \frac{1}{\cancel{63}} \cdot \cancel{63}$
$t_{63} = 1$

Answer 2

$\huge \bf Solution :$

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

$\bf T_{7} = \dfrac{1}{9}$

$\bf \implies a + 6d = \dfrac{1}{9} - ①$

$\bf T_{9} = \dfrac{1}{7}$

$\bf \implies a + 8d = \dfrac{1}{7} - ②$

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

$\bf (a + 8d) - (a + 6d) = \dfrac{1}{7} - \dfrac{1}{9}$

$\bf a + 8d - a - 6d = \dfrac{9}{63} - \dfrac{7}{63}$

$\bf 2d = \dfrac{9 - 7}{63}$

$\bf 2d = \dfrac{2}{63}$

$\bf d = \dfrac{2}{63 \times 2}$

$\bf d = \dfrac{\cancel{2}}{63 \times \cancel{2}}$

$\bf d = \dfrac{1}{63}$

$\large \boxed{\bf d = \dfrac{1}{63}}$

$\bf By, \: putting \: d = \dfrac{1}{63} \: in \: ①,$

$\bf we \: get$

$\bf a + 6 \times \dfrac{1}{63} = \dfrac{1}{9}$

$\bf a + \cancel{6}^{2} \times \dfrac{1}{\cancel{63}_{21}} = \dfrac{1}{9}$

$\bf a + \dfrac{2}{21} = \dfrac{1}{9}$

$\bf a = \dfrac{1}{9} - \dfrac{2}{21}$

$\bf a = \dfrac{7}{63} - \dfrac{6}{63}$

$\bf a = \dfrac{7 - 6}{63}$

$\bf a = \dfrac{1}{63}$

$\large \boxed{\bf a = \dfrac{1}{63}}$

$\bf Thus, \: a = \dfrac{1}{63} \: and \: d = \dfrac{1}{63}$

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

$\bf \implies T_{63} = \dfrac{1}{63} + 62 \times \dfrac{1}{63}$

$\bf \implies T_{63} = \dfrac{1}{63} + \dfrac{62}{63}$

$\bf \implies T_{63} = \dfrac{1 + 62}{63}$

$\bf \implies T_{63} = \dfrac{63}{63}$

$\bf \implies T_{63} = \cancel{\dfrac{63}{63}}$

$\bf \implies T_{63} = 1$

$\large \boxed{\bf T_{63} = 1 }$

$\bf Hence, \: 63rd \: term \: of \: given \: AP \: is \: 1.$