Question

Let $$T_r$$ be the $$r^{th}$$ term of an A.P., for $$r=1, 2, 3,$$_______. If for some positive integers m, n we have $$T_m=\dfrac{1}{n}$$ and $$T_n=\dfrac{1}{m}$$, then $$T_{mn}$$ equals.

A
1mn
B
1m+1n
C
1
D
0

Solution

The correct option is B $$1$$ Let, common difference$$=d$$ $${{r}^{th}}$$ term $$={{t}_{r}}$$ $${{1}^{st}}$$ term= a Now $${{m}^{th}}$$ term= $${{t}_{m}}=a+\left( m-1 \right)d$$ Since, $${{t}_{m}}=\dfrac{1}{n}$$ Therefore, $$\dfrac{1}{n}=a+\left( m-1 \right)$$      …… (1) Now $${{n}^{th}}$$ term $${{t}_{n}}=a+\left( m-1 \right)d$$ Given ,$${{t}_{n}}=\dfrac{1}{m}$$ Therefore, $$\dfrac{1}{m}=a+\left( n-1 \right)$$      …… (2) Subtract equation $$\left( 2 \right)$$ from equation$$\left( 1 \right)$$   $$\dfrac{1}{n}-\dfrac{1}{m}=\left( m-1 \right)d-\left( n-1 \right)d$$  $$\dfrac{m-n}{mn}=\left( m-n \right)d$$  $$d=\dfrac{1}{mn}$$ Put value d in equation $$\left( 1 \right)$$   $$\dfrac{1}{n}=a+\left( m-1 \right)\dfrac{1}{mn}$$  $$\dfrac{1}{n}=a+\dfrac{m}{mn}+\dfrac{1}{mn}$$  $$\dfrac{1}{n}=a+\dfrac{1}{n}-\dfrac{1}{mn}$$  $$\dfrac{1}{n}=a+\dfrac{1}{n}-\dfrac{1}{mn}$$  $$a=\dfrac{1}{mn}$$ Now $$m{{n}^{th}}$$ term ,   $${{t}_{mn}}=\dfrac{1}{mn}+\left( mn-1 \right)\dfrac{1}{mn}$$  $$=\dfrac{1}{mn}+1-\dfrac{1}{mn}$$  $${{t}_{mn}}=1$$ Mathematics

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