Question

If pth,qth and rth terms of an A.P. are a,b, c respectively, then show that a(q - r)+b(r - p)+c(p - q) = 0

Solve above question in a book or paper neatly to be marked as BRAINLIEST.
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Answer 1

$\huge{\text{\underline{Solution:-}}}$

★ Given:-

• Let a = First term of the AP
• Let d = Common difference of the AP

★ Now

• a = A + (p - 1) × d. → (1)
• b = A + (q - 1) × d → (2)
• c = A + (r - 1) × d → (3)

★ Point to remember:-

Subtracting 2nd from 1st , 3rd from 2nd and 1st from 3rd we get

• a - b = (p - q) × d → (4)
• b - c = (q - r) × d → (5)
• c - b = (r - p) × d → (6)

Multiply 4,5,6 by c,a,b respectively we have

• c × (a - b) = c × (p - q) × d = → 4)
• a × (b - c) = a × (q - r) × d = → (5)
• b × (c - a) = b × (r - p) × d = → (6)

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

Now since d is common difference it should be non zero

★ Therefore,

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

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Answer 2

Answer:

Refer the attached picture.