It tm = n and tn = m, show t(m+n) = 0
Answers
Answer:
t(m+n)= 0
Step-by-step explanation:
tm= n
∴n= a+(m-1)d....(1)
tn= m
∴m= a+(n-1)d....(2)
(1)-(2)⇒
∴n-m= a-a+md-d-nd+d
∴n-m= (m-n)d
∴d= (n-m)/(m-n)
Multiply the numerator and the denominator by (-1)
∴d= (n-m)(-1)/(m-n)(-1)
∴d= (n-m)(-1)/(n-m)
∴d= -1....(3)
Putting (3) in (1)
∴n= a+(m-1)(-1)
∴n= a+1-m
∴m+n= a+1
∴t(m+n)= t(a+1)
∴t(a+1)= a+(a+1-1)d
∴t(a+1)= a+a(-1)
∴t(a+1)= a-a
∴t(a+1)= 0
∴t(m+n)= 0 (∵t(m+n)= t(a+1))
Answer:
t(m+n)= 0
Step-by-step explanation:
tm= n
∴n= a+(m-1)d....(1)
tn= m
∴m= a+(n-1)d....(2)
(1)-(2)⇒
∴n-m= a-a+md-d-nd+d
∴n-m= (m-n)d
∴d= (n-m)/(m-n)
Multiply the numerator and the denominator by (-1)
∴d= (n-m)(-1)/(m-n)(-1)
∴d= (n-m)(-1)/(n-m)
∴d= -1....(3)
Putting (3) in (1)
∴n= a+(m-1)(-1)
∴n= a+1-m
∴m+n= a+1
∴t(m+n)= t(a+1)
∴t(a+1)= a+(a+1-1)d
∴t(a+1)= a+a(-1)
∴t(a+1)= a-a
∴t(a+1)= 0
∴t(m+n)= 0 (∵t(m+n)= t(a+1))