Question

if the 7th term of an ap is 1/9 and its 9th term is 1/7 find its 63rd term

Answer 1

nth term = a+ ( n-1 ) d
( where a is the first term, n is the no. of terms
and d is the difference between two consecutive terms . )

seventh term = a7 = 1/9
ninth term = a9 =1/7
a7 = a + 6d - ( 1 )
a9= a+ 8d - ( 2 )
by subtracting ( 1 ) from ( 2 )
(+ )a+ 8d = 1/7 ( +)
( - )a+ 6d = 1/9 ( - )
= 2d = 2 /63
d = 1/63
by putting value of d in eq. ( 2 )
a + 8 ( 1/63 ) = 1/7
a + 8/63 = 1/7
a = 1/7 - 8/63
a = 1/63

a 63 = a + 62d
=1/63 + 62 ( 1 /63 )
= 1/63 + 62/63
= 1 ans.

Answer 2

$\huge \bf Solution :$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf we \: get$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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