Question

7th term of an A. p is 1/9 and 9th term is 1/7.Find the 63rd term of the A. P.​

Answer 1

$\huge \bf Solution :$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf we \: get$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

___________________

Answer 2

Answer:

$\huge\fcolorbox{black}{lime}{❥ᴀ᭄ɴsᴡᴇʀ}$

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

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

7

=

9

1

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

9

1

−①

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

9

=

7

1

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

7

1

−②

\bf On \: subtracting \: ① \: from \: ②, \: we \: getOnsubtracting①from②,weget

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

7

1

−

9

1

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

63

9

−

63

7

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

63

9−7

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

63

2

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

63×2

2

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

63×

2

2

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

63

1

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

d=

63

1

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

63

1

in①,

\bf we \: getweget

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

63

1

=

9

1

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

6

2

×

63

21

1

=

9

1

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

21

2

=

9

1

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

9

1

−

21

2

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

63

7

−

63

6

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

63

7−6

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

63

1

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

a=

63

1

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

63

1

andd=

63

1

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

63

=a+(63−1)d=a+62d

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

63

=

63

1

+62×

63

1

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

63

=

63

1

+

63

62

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

63

=

63

1+62

\bf \implies T_{63} = \dfrac{63}{63}⟹T

63

=

63

63

\bf \implies T_{63} = \cancel{\dfrac{63}{63}}⟹T

63

=

63

63

\bf \implies T_{63} = 1⟹T

63

=1

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

T

63

=1

\bf Hence, \: 63rd \: term \: of \: given \: AP \: is \: 1.Hence,63rdtermofgivenAPis1.