Solve this


If $a, b, c$ are the $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of a GP, show that

$(q-r) \log a+(r-p) \log b+(p-q) \log c=0$


As per the question, $a, b$ and $c$ are the $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ term of GP.

Let us assume the required $\mathrm{GP}$ as $\mathrm{A}, \mathrm{AR}, \mathrm{AR}^{2}, \mathrm{AR}^{3} \ldots$

Now, the $\mathrm{n}^{\text {th }}$ term in the GP, $\mathrm{a}_{\mathrm{n}}=\mathrm{AR}^{\mathrm{n}-1}$

$p^{\text {th }}$ term, $a p=A R^{p-1}=a \rightarrow(1)$

$q^{\text {th }}$ term, $a_{q}=A R^{q-1}=b \rightarrow(2)$

$\mathrm{r}^{\text {th }}$ term, $\mathrm{a}_{\mathrm{r}}=\mathrm{AR}^{\mathrm{r}-1}=\mathrm{c} \rightarrow(3)$

$\frac{(1)}{(2)} \rightarrow \frac{\mathrm{R}^{\mathrm{p}-1}}{\mathrm{R}^{\mathrm{q}-1}}=\mathrm{R}^{\mathrm{p}-\mathrm{q}}=\frac{\mathrm{a}}{\mathrm{b}} \rightarrow$ (i)

$\frac{(2)}{(3)} \rightarrow \frac{\mathrm{R}^{\mathrm{q}-1}}{\mathrm{R}^{\mathrm{r}-1}}=\mathrm{R}^{\mathrm{q}-\mathrm{r}}=\frac{\mathrm{b}}{\mathrm{c}} \rightarrow$ (ii)

$\frac{(3)}{(1)} \rightarrow \frac{\mathrm{R}^{\mathrm{r}-1}}{\mathrm{R}^{\mathrm{p}-1}}=\mathrm{R}^{\mathrm{r}-\mathrm{p}}=\frac{\mathrm{c}}{\mathrm{a}} \rightarrow$ (iii)

Taking logarithm on both sides of equation (i), (ii) and (iii)

$(p-q) \log R=\log a-\log b$

$\therefore(\mathrm{p}-\mathrm{q})=\frac{\log \mathrm{a}-\log \mathrm{b}}{\log \mathrm{R}} \rightarrow(4)$

$(q-r) \log R=\log b-\log c$

$\therefore(q-r)=\frac{\log b-\log c}{\log R} \rightarrow(5)$

$(r-p) \log R=\log c-\log a$

$\therefore(\mathrm{r}-\mathrm{p})=\frac{\log \mathrm{c}-\log \mathrm{a}}{\log \mathrm{R}} \rightarrow(6)$

Now, multiply equation (4) with log c,

$(p-q) \log c=\left(\frac{\log a-\log b}{\log R}\right) \log c \rightarrow(7)$

Now, multiply equation (5) with log a,

$(\mathrm{q}-\mathrm{r}) \log \mathrm{a}=\left(\frac{\log \mathrm{b}-\log \mathrm{c}}{\log \mathrm{R}}\right) \log \mathrm{a} \rightarrow(8)$

Now, multiply equation (6) with log b,

$(r-p) \log b=\left(\frac{\log c-\log a}{\log R}\right) \log b \rightarrow(9)$

Now, add equations $(7),(8)$ and $(9)$.

$(p-q) \log c+(q-r) \log a+(r-p) \log b=\left(\frac{\log a-\log b}{\log R}\right) \log c$

$+\left(\frac{\log \mathrm{b}-\log \mathrm{c}}{\log \mathrm{R}}\right) \log \mathrm{a}+\left(\frac{\log \mathrm{c}-\log \mathrm{a}}{\log \mathrm{R}}\right) \operatorname{logb}$

On solving the above equation, we will get,

$(p-q) \log c+(q-r) \log a+(r-p) \log b=0$

Hence proved.


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