# 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 0 votes 23 views in Aptitude closed 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)

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