Question

koi solution bta do is question ka ​

Answer 1

Step-by-step explanation:

It is given that

a = 1/A (1)

a + d =  (2)

a + 2d = 1/C (3)

(1) IN (3)

1/A + 2d = 1/C

2d = 1/C - 1/A

2d = A - C / AC

d = A - C / 2AC

Therefore the the difference is A - C / 2AC . [because 1/A , 1/B , 1/C are in an AP , so the difference will be the same].

Answer 2

Answer:

Answer:

• Common difference,
• $d = \frac{a - c}{2ac}$

Given:

• Since, 1/a, 1/b ,1/c are in AP.

To prove:

• $d = \frac{a - c}{2ac}$

Solution:

As we know that,

The condition in AP is,

$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$

$\frac{2}{b} = \frac{a + c}{ac}$

$b = \frac{2ac}{a + c}$

Now,

Common difference,

$d = a2 - a1$

$d = \frac{1}{b} - \frac{1}{c}$

since,b=2ac/a+c

$d = \frac{a + c}{2ac} - \frac{1}{a}$

$d = \frac{a + c - 2c}{2ac}$

$d = \frac{a - c}{2ac}$

So, The Common difference,

$d = \frac{a - c}{2ac}$