Question

If the seventh term of an A.P. is $\frac{1}{9}$ and its ninth term is $\frac{1}{7}$, find its (63)^rd term.

Answer 1

Let a be the first term and d be the common difference.

We know that, n^th term = a_n = a + (n − 1)d

According to the question,

a₇ = $\frac{1}{9}$
⇒ a + (7 − 1)d = $\frac{1}{9}$
⇒ a + 6d = $\frac{1}{9}$ .... (1)

Also, a₉ = $\frac{1}{7}$
⇒ a + (9 − 1)d = $\frac{1}{7}$
⇒ a + 8d = $\frac{1}{7}$ ....(2)

On Subtracting (1) from (2), we get
8d − 6d = $\frac{1}{7}-\frac{1}{9}$
⇒ 2d = $\frac{9-7}{63}$
⇒ 2d = $\frac{2}{63}$
⇒ d = $\frac{1}{63}$
⇒ a = $\frac{1}{9}-\frac{6}{63}$ [From (1)]
⇒ a = $\frac{7-6}{63}$
⇒ a = $\frac{1}{63}$

∴ a₆₃ = a + (63 − 1)d
= $\frac{1}{63}+\frac{62}{63}$
= $\frac{63}{63}$
= 1

Thus, (63)^rd term of the given A.P. is 1.