If l, m and n are the pth, qth and rth terms of a G.P. and are all positive, then log⁡lp1log⁡mq1log⁡nr1equals

i=ARp−1⇒log⁡i=log⁡A+(p−1)log⁡R ⇒m=ARq−1⇒log⁡m=log⁡A+(q−1)log⁡R⇒n=ARr−1⇒log⁡n=log⁡A+(r−1)log⁡RNow, =log⁡l p 1log⁡m q 1log⁡n r 1 =log⁡A+(p−1)log⁡R p 1log⁡A+(q−1)log⁡R q 1log⁡A+(r−1)log⁡R r 1=log⁡A−log⁡R p 1log⁡A−log⁡R q 1log⁡A−log⁡R r 1+plog⁡R p 1qlog⁡R q 1rlog⁡R r 1 =(log⁡A−log⁡R)1p11q11r1+log⁡Rpp1qq1rr1=0

If l, m and n are the pth, qth and rth terms of a G.P. and are all positive, then log⁡lp1log⁡mq1log⁡nr1equals

i=ARp−1⇒log⁡i=log⁡A+(p−1)log⁡R ⇒m=ARq−1⇒log⁡m=log⁡A+(q−1)log⁡R⇒n=ARr−1⇒log⁡n=log⁡A+(r−1)log⁡RNow, =log⁡l p 1log⁡m q 1log⁡n r 1 =log⁡A+(p−1)log⁡R p 1log⁡A+(q−1)log⁡R q 1log⁡A+(r−1)log⁡R r 1=log⁡A−log⁡R p 1log⁡A−log⁡R q 1log⁡A−log⁡R r 1+plog⁡R p 1qlog⁡R q 1rlog⁡R r 1 =(log⁡A−log⁡R)1p11q11r1+log⁡Rpp1qq1rr1=0

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If l, m and n are the p^th, q^thand r^th terms of a G.P. and are all positive, then $[\begin{array}{ccc} \log l & p & 1 \\ \log m & q & 1 \\ \log n & r & 1 \end{array}]$ equals

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Detailed Solution

$\begin{array}{l} i = {AR}^{p - 1} \Rightarrow \log i = \log A + (p - 1) \log R \Rightarrow m = {AR}^{q - 1} \\ \Rightarrow \log m = \log A + (q - 1) \log R \\ \Rightarrow n = {AR}^{r - 1} \Rightarrow \log n = \log A + (r - 1) \log R \end{array}$
Now,

$= |\begin{array}{lll} \log l p 1 \\ \log m q 1 \\ \log n r 1 \end{array}| = |\begin{array}{lll} \log A + (p - 1) \log R p 1 \\ \log A + (q - 1) \log R q 1 \\ \log A + (r - 1) \log R r 1 \end{array}|$

$= |\begin{array}{lll} \log A - \log R p 1 \\ \log A - \log R q 1 \\ \log A - \log R r 1 \end{array}| + |\begin{array}{lll} plog R p 1 \\ qlog R q 1 \\ rlog R r 1 \end{array}|$
$= (\log A - \log R) |\begin{array}{ccc} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \end{array}| + \log R |\begin{array}{ccc} p & p & 1 \\ q & q & 1 \\ r & r & 1 \end{array}|$