Let Tr be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, m â n, Tm = (1/n) and Tn = (1/m), then a - d equals

Question

Let Tr be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, m â n, Tm = (1/n) and Tn = (1/m), then a - d equals (A) 0 (B) 1 (C) (1/mn) (D) (1/m) + (1/n)

Tardigrade Admin · Accepted Answer

Tm = (1/n) = a+(m-1)d .....(1) and Tn = (1/m) = a+(n-1)d .....(2) from (1) and (2) we get a = (1/mn), d = (1/mn) Hence a - d = 0

Q. Let $T_{r}$ be the rth term of an $A . P$ . whose first term is a and common difference is d. If for some positive integers $m, n, m \neq = n,$ $T_{m} = \frac{1}{n}$ and $T_{n} = \frac{1}{m},$ then $a - d$ equals

$0$

$1$

$\frac{1}{mn}$

$\frac{1}{m} + \frac{1}{n}$

$T_{m} = \frac{1}{n} = a + (m - 1) d ..... (1)$
and $T_{n} = \frac{1}{m} = a + (n - 1) d ..... (2)$
from $(1)$ and $(2)$ we get $a = \frac{1}{mn}, d = \frac{1}{mn}$
Hence $a - d = 0$

Q. Let Tr​ be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m,n,m=n, Tm​=n1​ and Tn​=m1​, then a−d equals

0

1

mn1​

m1​+n1​

Tm​=n1​=a+(m−1)d.....(1) and Tn​=m1​=a+(n−1)d.....(2) from (1) and (2) we get a=mn1​,d=mn1​ Hence a−d=0

Q. Let $T_{r}$ be the rth term of an $A . P$ . whose first term is a and common difference is d. If for some positive integers $m, n, m \neq = n,$ $T_{m} = \frac{1}{n}$ and $T_{n} = \frac{1}{m},$ then $a - d$ equals

$0$

$1$

$\frac{1}{mn}$

$\frac{1}{m} + \frac{1}{n}$

$T_{m} = \frac{1}{n} = a + (m - 1) d ..... (1)$
and $T_{n} = \frac{1}{m} = a + (n - 1) d ..... (2)$
from $(1)$ and $(2)$ we get $a = \frac{1}{mn}, d = \frac{1}{mn}$
Hence $a - d = 0$