19. If the pg", and "terms of an A. P. are a, b, and c respectively. prove that: p(b-c) + (ca) + rſa-b) = 0 50 Approved by Curriculum Development Centre​

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

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

0.85% : )

step-by-step explanation: