Answer:
Step-by-step explanation:a = m + (p-1)*d
b = m + (q-1)*d
c = m + (r-1)*d
p(b-c) = p*(q-r)*d
q(c-a) = q*(r-p)*d
r(a-b) = r*(p-q)*d
p(b-c)+q(c-a)+r(a-b)
= p*(q-r)*d + q*(r-p)*d +r*(p-q)*d
= (pq-pr+qr-pq+rp-qr)*d
= 0*d = 0
So i prove p(b-c)+q(c-a)+r(a-b)=0 hope this is helpfull
Answer:
3
Step-by-step explanation:
as a person who answers in the mathematics session
i will tell you that the answer is 3
Three hundred thousand four thousand one
300000 +4000+1
