Question

If the 9th term of an ap is 1/7 and the 7th term os 1/9 find the 63rd term

Answer 1

Answer:

$a_{63}$ = 1

Step-by-step explanation:

The n th term of an arithmetic progression is

$a_{n}$ = a + (n - 1)d

where a is the first term and d the common difference

use the 9 th term and 7 th term to find a and d

$a_{9}$ = a + 8d = $\frac{1}{7}$ → (1)

$a_{7}$ = a + 6d = $\frac{1}{9}$ → (2)

Subtract (2) from (1) term by term

2d = $\frac{2}{63}$ ⇒ d = $\frac{1}{63}$

Substitute this value into (2) and solve for a

a + $\frac{6}{63}$ = $\frac{1}{9}$

a = $\frac{1}{9}$ - $\frac{6}{63}$ = $\frac{1}{63}$

Hence

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