Question

If the seventh term of an A.P. is `(1)/(9)` and its ninth term is `(1)/(7)`, find its 63rd term.

Answer 1

a₇ = `(1)/(9)`
⇒ a + 6d = `(1)/(9)`….(i)
a₉ = `(1)/(7)`
⇒ a + 8d = `(1)/(7)`……(ii)
a₇ = `(1)/(9) ⇒ a + 6d = (1)/(9)`        ....(i)

a₉ = `(1)/(7) ⇒ a + 8d = (1)/(7)`        ....(i)
                    –      –          –    
On subtracting, –2d`(1)/(9) - (1)/(7)`

–2d = `(7 - 9)/(63)`

–2d = `(-2)/(63)`

d = `(1)/(63)`
Now, substitute the value of d in eq. (i), we get

`a + 6(1/63) = (1)/(9)`

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

= `(7 - 6)/(63)`

= `(1)/(63)`
∴ a₆₃ = a + 62d

= `(1)/(63) + 62(1/63)`

= `(1 + 62)/(63)`

= `(63)/(63)`
= 1.