Question

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

Answer 1

here both a and d will be same, then we will get the value of a63 = 1

Answer 2

7th term = 1 / 9

a + ( 7 - 1 ) d = 1 / 9

a + 6d = 1 / 9

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

9th term = 1 / 7

a + ( 9 - 1 ) d = 1 / 7

a + 8d = 1 / 7

a = 1 / 7 - 8d             ------- : ( 2 )

Comparing both the equations :

1 / 9 - 6d = 1 / 7 - 8d

8d - 6d = 1 / 7 - 1 / 9

2d = ( 9 - 7 ) / 63

2d = 2 / 63

d = 1 / 63

  Putting the value of d in ( 1 ) :

a = 1 / 9 - 6d

   = 1 / 9 - 6( 1 / 63 )

   = ( 7 - 6 )  / 63

   = 1 / 63

Now, a = 1 / 63 and d = 1 / 63

Therefore,

$a_{63} = a + ( 63 - 1 )d$

$a_{63} = \dfrac{1}{63} + ( 62 \times \dfrac{1}{63} )$

$a_{63} = \dfrac{1 + 62}{63}$

$a_{63} = \dfrac{63}{63}$

$a_{63} = 1$

Therefore, 63rd term of the AP is 1.