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Let Tr, be the rth term of an A. P. for r = 1, 2, 3, _____ If for some positive integers m and n, we have \(T_m=\dfrac{1}{n}\:\text{and}\:T_n=\dfrac{1}{m},\) then Tmn equals:

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Correct Answer - Option 4 : 1

Given:

Series is AP (Arithmetic Progression)

mth Term = Tm = \(\dfrac{1}{n}\)

nth Term = Tn = \(\dfrac{1}{m}\)

Concept:

Arithmetic Progression Formulas

For the series, a1, a2, a3, a4....., an

an = a1 + (n - 1) d

where,

an = nth term of the series

a1 = First term of the series

n = number of term

d = Common Difference

Solution:

Let,

First Term = a

rth Term = Tr

Common Difference = d

So, mth Term = T= \(\dfrac{1}{n}\) =

nth Term = T= \(\dfrac{1}{m}\) =

Putting the value of d in (1)

\(\dfrac{1}{n}\) =

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

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

a = \(\dfrac{1}{mn}\)

For mnth term,

Tmn = a + (mn - 1)d

Tmn\(\dfrac{1}{mn}\) + (mn - 1) × \(\dfrac{1}{mn}\)

Hence, we get Tmn = 1


For AP (Arithmetic Progression):

an = a1 + (n - 1)d

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

For GP (Geometric Progression):

an = a1 × r(n - 1)

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

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