Question

if pth qth rth term of AP are a b c respectively then show that into cube minus or plus SBI into a minus b + c into P - Q is equal

Answer 1

Let the first term of the AP = A

let the common difference = D

For an AP, we know that

Tn = A + (n-1)D, so

pth term = a = A + (p-1)D

=> a(q-r) = A(q-r) + (q-r)(p-1)D................eqn1

qth term = b = A + (q-1)D

=> b(r-p) = A(r-p) + (r-p)(q-1)D.............................eqn2

rth term = c = A + (r-1)D

=> c(p-q) = A(p-q) + (p-q)(r-1)D.....................eqn3

Adding eqn1 eqn2 and eqn3 we get

a(q-r)+b(r-p) + c(p-q) = A(q-r) + (q-r)(p-1)D + A(r-p)+(r-p)(q-1)D+A(p-q)+(p-q)(r-1)D

= A(q-r + r-p + p-q) + D[(q-r)(p-1) + (r-p)(q-1) + (p-q)(r-1)]

= A(0) + D[ pq - q - pr + r + qr - r - pq + p + pr - p - qr + q]

= 0 + D(0)

= 0

Hence,

a(q-r)+b(r-p) + c(p-q) = 0

Proved.

Answer 2

Step-by-step explanation:

answer is zero

we should multiply