Question

If the 7th term of ap is 1/9 and 9th term is 1/7 then 63rd term is

Answer 1

a₇ =1/9 ; a₉=1/7

a+8d-a - 6d = 1/7 - 1/9

2d = (9-7)/35 = 2/35

d= 2/35*2

d=1/35

a+6d = 1/9

a = 1/9 - 6/35 = 1/35

a₆₃ = a+(n-1)d = 1/35 +(63 - 1)1/35 = (1 + 62)/35 = 63/35 = 9/5

Answer 2

Answer:

The 63rd term of the AP is 1.

Step-by-step explanation:

Given to us:

7th term of the AP = 1/9

9th term of the AP = 1/7

The formula used to find the terms are:

${\boxed{\boxed{\sf{\implies \an = a + (n - 1)d}}}}$

Using this formula we frame equations:

a7 = a + (7 - 1)d

a7 = a + 6d

a + 6d = 1/9  ---------------------(Eq - 1)

a + 8d = 1/7  --------------------(Eq - 2)

Solving Eq 1 and Eq 2 by Elimination method:

a + 6d = 1/9  

a + 8d = 1/7

--------------------

   - 2d = 7 - 9 / 63 (Rationalizing denominators)

   - 2d = - 2 / 63

   d = - 2 / 63 * - 1 / 2

   d = 1 / 63

We know that,

a + 6d = 1/ 9

Substituting value of a and d in this equation:

a + 6 (1/63) = 1/9

a + 6/63 = 1/9

a = 1/9 - 6/63

a = 7 - 6 / 63 (Rationalizing the denominators)

a = 1/63

Now,

a + 62d = 63rd term of the AP

Substituting the value of a and d in the following equation,

=> 1/63 + 62(1/63)

=> 1/63 + 62/63

=> 1 + 62/63

=> 63/63

=> 1

Therefore, the 63rd term of the AP is 1.