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Exercise 5.4 · Q10

Q.Three positive numbers form an increasing G.P. If the middle term of the series is doubled, then the new numbers are in A.P. Find the common ratio of the G.P.

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Represent the increasing GP as a,ar,ar2a,ar,ar^2, apply the AP condition to a,2ar,ar2a,2ar,ar^2, solve the resulting quadratic in rr, and reject the root that is not consistent with an increasing GP.

Three numbers p,q,sp,q,s are in AP iff 2q=p+s2q=p+s. A GP with common ratio rr is increasing (for positive terms) iff r>1r>1.

  1. Let the three positive numbers in increasing GP be a, ar, ar2a,\ ar,\ ar^2 with a>0a>0 and r>1r>1.
  2. Doubling the middle term gives the new numbers a, 2ar, ar2a,\ 2ar,\ ar^2, which are stated to be in AP.
  3. Apply the AP condition (twice the middle term equals the sum of the outer terms):

2(2ar)=a+ar22(2ar)=a+ar^2

  1. Simplify:

4ar=a+ar24ar=a+ar^2

  1. Divide throughout by aa (since a>0a>0, a≠0a\neq0):

4r=1+r24r=1+r^2

  1. Rearrange into standard quadratic form:

r2−4r+1=0r^2-4r+1=0

  1. Solve using the quadratic formula r=−b±b2−4ac2ar=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=1,b=−4,c=1a=1,b=-4,c=1: …

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