Question

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

Answer 1

Morning Yarr Answer;-)


T7 = 1/9
a+6d = 1/9 
a = (1/9)-6d -------------(1)

T9 = 1/7
a+8d = 1/7
a = (1/7)-8d --------------(2)

(1/9)-6d = (1/7)-8d
8d-6d = (1/7)-(1/9)
2d = 2/63

d = 1/63

a = (1/9)-6d
a = (1/9) - (6/63)
a = 1/63

T63
= a+62d
= (1/63) + 62×(1/63)
= (1/63)+(62/63)
= 63/63
= 1
====================
Hence the 63rd term of the given A.P is 1

Thnx for ask this question...

Answer 2

$\huge \bf Solution :$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf we \: get$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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