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Home » If a, b, c, d are in GP, prove that If a, b, c, d are in GP, prove that (i) (b + c)(b + d) = (c + a)(c + a) (ii) (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2
If a, b, c, d are in GP, prove that If a, b, c, d are in GP, prove that
(i) (b + c)(b + d) = (c + a)(c + a)
(ii) (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2
CBSE | Class 11 | Excercise 12E | Geometrical Progression
If a, b, c, d are in GP, prove that If a, b, c, d are in GP, prove that
(i) (b + c)(b + d) = (c + a)(c + a)
(ii) (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2

Hence Proved

(iii) (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2

To prove: (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2 Given: a, b, c, d are in GP

Proof: When a,b,c,d are in GP thenFrom the above, we can have the following conclusion

⇒ bc = ad … (i)

⇒ b2 = ac … (ii)

⇒ c2 = bd … (iii)

Taking LHS = (a + b + c + d)2

⇒ (a + b + c + d) (a + b + c + d)

⇒ a2 + ab + ac + ad + ba + b2 + bc + bd + ca + cb + c2 + cd + da + db + dc + d2 On rearranging

⇒ a2 + ab + ba +b2 + ac + ad + bc + bd + ca + cb + c2 + cd + da + db + dc + d2 On rearranging

⇒ (a + b)2 + ac + ad + bc + bd + ca + cb + da + db + c2 + cd + dc + d2 On rearranging

⇒ (a + b)2 + ac + ad + bc + bd + ca + cb + da + db + (c + d)2 On rearranging

⇒ (a + b)2 + ac + ca + ad + bc + cb + da + bd + db + (c + d)2 Using eqn. (i)

⇒ (a + b)2 + ac + ca + bc + bc + bc + bc + bd + db + (c + d)2 Using eqn. (ii)

⇒ (a + b)2 + b2 + b2 + bc + bc + bc + bc + bd + db + (c + d)2 Using eqn. (iii)

⇒ (a + b)2 + 2b2 + 4bc + c2 + c2 + (c + d)2 On rearranging

⇒ (a + b)2 + 2b2 + 4bc + 2c2 + (c + d)2

⇒ (a + b)2 + 2[b2 + 2bc + c2 ] + (c + d)2

⇒ (a + b)2 + 2(b + c)2 + (c + d)2

= RHS

Hence proved

 

i

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