Question

if the seventh term of an AP is 1/9and it's ninth term is 1/7so find its 63rd term

Answer 1

let a is the first term and d is the common difference
now ,
7th term =1/9
a +(7- 1) d = 1/9 {use Tn =a+(n-1) d
a +6d =1/9 -----------------(1)

again ,

9th term = 1/7
a +(9- 1) d = 1/7
a + 8 d = 1/7 --------------(2)

subtracting equation (2) to (1)
2d =1/7 - 1/9 = 2/63

d= 1/63
put d = 1/63 in equation (1)
a= 1/9 - 6/63 = 1/9 - 2/21

a = (21-18)/21 x 9

=3 /21 x 9

=1 /63
now ,
63th term = a+ (63-1) d
=1 /63 + 62/63 = 63/63 = 1

hence 63th term =1

Answer 2

we know that nth term of an A.P is given as...
a + (n -1)d = to
let a be the 1st term and d be the common difference

so it's seventh term

t7 = 1/9

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

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

and it's ninth term is ...

t9 = 1/7

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

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

now subtracting (2) - (1) we get...

a+ 8d = 1/7
a + 6d= 1/9
-. -. -.
------------------------
0 + 2d = 1 / 7 - 1/ 9

={ 9 - 7 } /63

2d = 2/63

d = 1/63

so put d= 1/63 in equation ( 1) to find value of a...

a + 6 × 1/63 = 1/9

a = 1/9 - 6 /63

= { 7 - 6} /63 = 1/63

so it's 63rd term is

t63 = 1/63 + (63- 1) ×1/63

= 1/63 + 62/63

={ 1+ 62} / 63 = 63 / 63 = 1 ans...