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  • If pth, qth, and rth terms of an A.P. and G.P. are both a, b and c respectively, show that a^b–c . b^c – a. c^a – b = 1

If p^{th}, q^{th}, and r^{th} terms of an A.P. and G.P. are both a, b and c respectively, show that

a\textsuperscript{b-c} . b\textsuperscript{c - a}. c\textsuperscript{a - b} = 1

Answers (1)

Let the first term of AP be m and common difference as d. Let the first term of GP be I and common ratio be s.

For AP

\\ t_{p}=m+ \left( p-1 \right) d \\\\ a=m+ \left( p-1 \right) d \\\\ b=m+ \left( q-1 \right) d \\\\ c=m+ \left( r-1 \right) d \\\\

For GP

 


\\ t_{p}=Is^{p-1} \\\\ a=Is^{p-1} \\\\ b=Is^{q-1} \\\\ c=Is^{r-1} \\\\ b-c= \left( q-r \right) d \\\\


 \\ c-a= \left( r-p \right) d \\\\ a-b= \left( p-q \right) d \\\\ LHS=a^{b-c}b^{c-a}c^{a-b} \\\\ = \left( Is^{p-1} \right) ^{ \left( q-r \right) d} \left( Is^{q-1} \right) ^{ \left( r-p \right) d} \left( Is^{r-1} \right) ^{ \left( p-q \right) d} \\\\ =I^{ \left( q-r+r-p+p-q \right) d}s^{ \left[ \left( pq-pr-q+r \right) + \left( qr-qp-r+p \right) + \left( rp-rq-p+q \right) \right] d} \\\\ =I^{0}s^{0}=1 \\\\

 

 

Posted by

infoexpert21

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