Question

Let T_r, be the r^th 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 T^mn equals:

Answer 1

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, a₁, a₂, a₃, a₄....., a_n

a_n = a₁ + (n - 1) d

where,

a_n = n^th term of the series

a₁ = 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

⇒ 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 = a₁ + (n - 1)d

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

For GP (Geometric Progression):

a_n = a₁ × r^{(n - 1)}

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