If seventh term of AP is 1/9 and it 's ninth term is 1/7 find its 63rd

1.	If seventh term of AP is 1/9 and it 's ninth term is 1/7 find its 63rd term
Answer» $\huge{\bf{\underline{\purple{Solution :}}}}$ Given, Seventh Term of AP $\bold{a_{7} = \frac{1}{9}}$ Ninth Term $\bold{a_{9} = \frac{1}{7}}$ We KNOW that : $\boxed{\bold{\red{n^{th} \ term = a + (n - 1) d}}}$ Here , a is the first term . n is the number of TERMS . d is the difference between TWO Consecutive terms . $\bold{\pink{First \ Case :}}$ $\implies \bold{a_{n} = a + (n + 1) d}$ $\implies \bold{a_{7} = a + (7 -1) d}$ $\implies \bold{\frac{1}{9} = a + 6d}$ $\implies \boxed{\bold{a = \frac{1}{9} - 6d}}$ ........................1) Equation $\bold{\pink{Second \ Case :}}$ $\implies \bold{a_{n} = a + (n -1) d}$ $\implies \bold{a_{9} = a + (9 -1) d}$ $\implies \bold{\frac{1}{7} = a + 8d}$ $\implies \boxed{\bold{ a = \frac{1}{7} - 8d}}$ .....................2) Equation Now From Equation 1) and 2) :- $\implies \bold{\frac{1}{9} - 6d = \frac{1}{7} - 8d}$ $\implies \bold{ 8d -6d = \frac{1}{7} - \frac{1}{9}}$ $\implies \bold{2d = \frac{2}{63}}$ $\implies \boxed{\bold{ d = \frac{1}{63}}}$ Now Put the value of d in Equation 2) :- $\implies \bold{a = \frac{1}{7} - 8 ( \frac{1}{63})}$ $\implies \bold{ a = \frac{1}{7} - \frac{8}{63}}$ $\implies \boxed{\bold{ a = \frac{1}{63}}}$ Now Find 63RD term of AP :- $\implies \bold{a_{n} = (n - 1) d}$ $\implies \bold{a_{63} = (63 - 1) \frac{1}{63}}$ $\implies \bold{a_{63} = \frac{1}{63} + \frac{62}{63}}$ $\implies \huge{\boxed{\bold{\red{a_{63} = 1}}}}$ $\rule{200}2$