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Exercise 5.1 · Q16

Q.If ap+bpap−1+bp−1\dfrac{a^p+b^p}{a^{p-1}+b^{p-1}} is the A.M. between aa and bb, then find the value of pp. Also, for p,q,rp, q, r in A.P., prove that (a−b)r+(b−c)p+(c−a)q=0(a-b)r + (b-c)p + (c-a)q = 0. [Note: the book's expression appeared truncated at the page break; the identity is completed with "=0= 0" as strongly implied by the "prove that" phrasing and the standard result.]

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Setting the given expression equal to the A.M. of a,ba,b forces p=1p=1; separately, if a,b,ca,b,c are the ppth, qqth, rrth terms of an A.P., the stated identity reduces to 0=00=0.

A.M. of aa and bb: a+b2\dfrac{a+b}{2}. If a,b,ca,b,c are the ppth, qqth, rrth terms of an A.P. with first term AA and common difference DD: a=A+(p−1)D, b=A+(q−1)D, c=A+(r−1)Da=A+(p-1)D,\ b=A+(q-1)D,\ c=A+(r-1)D.

Note on the source text: the printed problem breaks across a page and the second identity's right-hand side is cut off; it is completed as "=0=0", the standard form of this well-known result, and is proved below under the standard reading that a,b,ca,b,c are respectively the ppth, qqth, rrth terms of an A.P. (which is exactly what makes the printed left-hand side identically zero).

Part 1 — finding pp:

  1. Given ap+bpap−1+bp−1=a+b2\dfrac{a^p+b^p}{a^{p-1}+b^{p-1}}=\dfrac{a+b}{2} (the A.M. of aa and bb).
  2. Cross-multiply: 2(ap+bp)=(a+b)(ap−1+bp−1)=ap+abp−1+ap−1b+bp2(a^p+b^p)=(a+b)(a^{p-1}+b^{p-1})=a^p+ab^{p-1}+a^{p-1}b+b^p.
  3. Simplify: 2ap+2bp−ap−bp=abp−1+ap−1b⇒ap+bp=abp−1+ap−1b2a^p+2b^p-a^p-b^p=ab^{p-1}+a^{p-1}b\Rightarrow a^p+b^p=ab^{p-1}+a^{p-1}b.
  4. Rearrange: ap−ap−1b=abp−1−bp⇒ap−1(a−b)=bp−1(a−b)a^p-a^{p-1}b=ab^{p-1}-b^p\Rightarrow a^{p-1}(a-b)=b^{p-1}(a-b).
  5. Since a≠ba\ne b, divide by (a−b)(a-b): ap−1=bp−1⇒(ab)p−1=1⇒p−1=0⇒p=1a^{p-1}=b^{p-1}\Rightarrow \left(\dfrac{a}{b}\right)^{p-1}=1\Rightarrow p-1=0\Rightarrow p=1. …

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