Question

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

Answer 1

Answer:

See The Attachment My Friend.....

Answer 2

$\huge{\bf{\underline{\purple{Solution :}}}}$

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 :-

$\implies \bold{a_{n} = (n - 1) d}$

$\implies \bold{a_{63} = (63 - 1) \frac{1}{63}}$

$\implies \bold{a_{63} = \frac{1}{63} + \frac{62}{63}}$

$\implies \huge{\boxed{\bold{\red{a_{63} = 1}}}}$

$\rule{200}2$