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Exercise 5.3 · Q5

Q.If a,ba, b and cc are in A.P. as well as in G.P., then which of the following is true?

(a) a=b≠ca=b \neq c
(b) a≠b≠ca \neq b \neq c
(c) a=b=ca=b=c
(d) a≠b=ca \neq b = c
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Numbers that are simultaneously in AP and GP must all be equal — option (c).

[!FORMULA] AP condition: 2b=a+c2b=a+c; GP condition: b2=acb^2=ac

a,b,ca,b,c = the three terms.

  1. Since a,b,ca,b,c are in AP: 2b=a+c2b=a+c, so c=2b−ac=2b-a.
  2. Since a,b,ca,b,c are also in GP: b2=acb^2=ac.
  3. Substitute c=2b−ac=2b-a into the GP condition: b2=a(2b−a)=2ab−a2b^2=a(2b-a)=2ab-a^2. …

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