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If a, b, c are in A.P., prove that: \begin{array}{l} (i){(a - c)^2} = 4(a - b)(b - c)\\ (ii){a^2} + {c^2} + 4ac = 2(ab + bc + ca) \end{array}
Arithmetic Progressions | CBSE | Class 11 | Exercise 19.5 | Maths | RD Sharma
If a, b, c are in A.P., prove that: \begin{array}{l} (i){(a - c)^2} = 4(a - b)(b - c)\\ (ii){a^2} + {c^2} + 4ac = 2(ab + bc + ca) \end{array}

Answer:

(i) 

Expanding,

a2 + c2 – 2ac = 4(ab – ac – b2 + bc)

a2 + 4c2b2 + 2ac – 4ab – 4bc = 0

(a + c – 2b)2 = 0

a + c – 2b = 0

a, b, c are in AP

b – a = c – b

a + c – 2b = 0

a + c = 2b

(a – c)2 = 4 (a – b) (b – c)

(ii) 

Expanding,

a2 + c2 + 4ac = 2 (ab + bc + ca)

a2 + c2 + 2ac – 2ab – 2bc = 0

(a + c – b)2 – b2 = 0

a + c – b = b

a + c – 2b = 0

2b = a+c

b = (a+c)/2

a, b, c are in AP

b – a = c – b

b = (a+c)/2

a2 + c2 + 4ac = 2 (ab + bc + ca)

i

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