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

Given,

a₇ = $\dfrac{1}{9}$     (Eq 1)

a₉ = $\dfrac{1}{7}$     (Eq 2)

By using formula a_n = a + (n - 1)d for Eq 1 we get,

⇒ a₇ = a + (7 - 1)d = $\dfrac{1}{9}$

⇒ a + 6d = $\dfrac{1}{9}$

⇒ 9(a + 6d) = 1

⇒ 9a + 54d = 1

⇒ 9a = 1 - 54d

⇒ a = $\dfrac{1 - 54d}{9}$     (Eq 3)

By using formula a_n = a + (n - 1)d for Eq 2 we get,

⇒ a₉ = a + (9 - 1)d = $\dfrac{1}{7}$

⇒ a + 8d = $\dfrac{1}{7}$     (Eq 4)

Putting value of a from Eq 3 in Eq 4 above,

$\Rightarrow \dfrac{1 - 54d}{9} + 8d = \dfrac{1}{7} \\[1em] \Rightarrow \dfrac{1 - 54d + 72d}{9} = \dfrac{1}{7} \\[1em] \Rightarrow \dfrac{1 + 18d}{9} = \dfrac{1}{7} \\[1em] \Rightarrow 7 + 126d = 9 \\[1em] \Rightarrow 126d = 2 \\[1em] \Rightarrow d = \dfrac{2}{126} \\[1em] \Rightarrow d = \dfrac{1}{63} \\[1.5em] \therefore a = \dfrac{1 - 54d}{9} \\[1em] = \dfrac{1 - 54 \times \dfrac{1}{63}}{9} \\[1em] = \dfrac{63 - 54}{9 \times 63} \\[1em] = \dfrac{9}{9 \times 63} \\[1em] = \dfrac{1}{63}. \\[1em]$

63rd term = a₆₃ = $a + (63 - 1)d$

$= a + 62d \\[1em] = \dfrac{1}{63} + 62 \times \dfrac{1}{63} \\[1em] = \dfrac{1}{63} + \dfrac{62}{63} \\[1em] = \dfrac{63}{63} \\[1em] = 1.$

Hence, 63rd term of the A.P. is 1.