Question

please solve this fastly it is very urgent ​

Answer 1

Step-by-step explanation:

Let the 1st term be a and common difference be d

Now, according to the Question,

a(mth) = (1/n)

a(nth) = (1/m)

Now,

we know that, to find the nth term we use the formula,

a(nth) = a + (n - 1)d

Hence,

a(mth) = a + (m - 1)d

Thus,

a + (m - 1)d = (1/n) ------ 1

Also,

a(nth) = a + (n - 1)d

Thus,

a + (n - 1)d = (1/m) ------- 2

Subtracting eq.1 and eq.2 we get,

a + (m - 1)d = (1/n)

- (a + (n - 1)d = (1/m))

(-) (-) (-)

——————————

0 + (m - 1)d - (n - 1)d = (1/n) - (1/m)

———————————

Thus,

(m - 1)d - (n - 1)d = (1/n) - (1/m)

Taking d common and LCM of m and n

d((m - 1) - (n - 1)) = (m - n)/mn

d(m - 1 - n + 1) = (m - n)/mn

d(m - n) = (m - n)/mn

d = ((m - n)/mn) × (1/(m - n))

d = (1/mn) ------ 3

Now,

Putting d = (1/mn) in eq.1

a + (m - 1)(1/mn) = (1/n)

Using Distributive Property,

a + (m × (1/mn) - 1 × (1/mn)) = (1/n)

a + (1/n) - (1/mn) = (1/n)

a = (1/n) - (1/n) + (1/mn)

a = (1/mn) ------ 4

Thus, now we can find what mnth term is,

a(mnth) = a + (mn - 1)d

From eq.3 and eq.4,

= (1/mn) + (mn - 1)(1/mn)

Using Distributive Property,

= (1/mn) + ((mn × (1/mn)) - (1 × (1/mn))

= (1/mn) + 1 - (1/mn)

= 1

Hence proved

Thus,

a(mnth) = 1

It might be a bit confusing and difficult to understand the typings, so please do refer the above image.

Hope it helped and believing you understood it........All the best