Correct Answer - Option 4 : 1

**Given:**

Series is AP (Arithmetic Progression)

m^{th} Term = T_{m} = \(\dfrac{1}{n}\)

n^{th} Term = T_{n} = \(\dfrac{1}{m}\)

**Concept:**

Arithmetic Progression Formulas

For the series, a_{1}, a_{2}, a_{3}, a_{4}....., a_{n}

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

where,

a_{n} = n^{th} term of the series

a_{1} = First term of the series

n = number of term

d = Common Difference

**Solution:**

Let,

First Term = a

r^{th} Term = T_{r}

Common Difference = d

So, m^{th} Term = T_{m }= \(\dfrac{1}{n}\) = $a+(m−1)d ---(1)$

$So,$n^{th} Term = T_{n }= \(\dfrac{1}{m}\) = $a+(n−1)d ---(2)$

$Subtracting (2) from (1),$

$\(\dfrac {1}{n}- \dfrac{1}{m} \)(m−1)d−(n−1)d$

$\(\dfrac{m-n}{mn}\)(m−n)d$

$\(\dfrac{1}{mn}\)$

Putting the value of **d** in **(1)**

⇒ \(\dfrac{1}{n}\) = $a+(m−1\(\dfrac{1}{mn}\)$

= \(\dfrac{1}{n}\) = a + \(\dfrac{m}{mn}\) - \(\dfrac{1}{mn}\)

= \(\dfrac{1}{n}\) = a + \(\dfrac{1}{n}\) - \(\dfrac{1}{mn}\)

⇒ **a **=** \(\dfrac{1}{mn}\)**

For mn^{th} term,

T_{mn} = a + (mn - 1)d

T_{mn} = \(\dfrac{1}{mn}\) + (mn - 1) × \(\dfrac{1}{mn}\)

Hence, we get **T**_{mn} = 1

For AP (Arithmetic Progression):

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

\(S_n = \dfrac{n}{2}[2 a_1+(n-1)d]\)

For GP (Geometric Progression):

a_{n} = a_{1} × r^{(n - 1)}

\(Sn = \dfrac{a_1(r^n - 1)}{(r-1)}\)