Question

If pth, qth , rth terms of a gp are a,b,c then ∑(q-r)log a is equal to

Answer 1

Dear student,

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{is}\mathrm{x}\mathrm{and}\mathrm{common}\mathrm{ratio}\mathrm{is}\mathrm{y} \\ \mathrm{a}={\mathrm{xy}}^{\mathrm{p}-1} \\ \mathrm{b}={\mathrm{xy}}^{\mathrm{q}-1} \\ \mathrm{c}={\mathrm{xy}}^{\mathrm{q}-1} \\ \mathrm{Now} \\ \sum _{}^{}\left(\mathrm{q}-\mathrm{r}\right)\mathrm{loga}=\left(\mathrm{q}-\mathrm{r}\right)\mathrm{loga}+\left(\mathrm{r}-\mathrm{p}\right)\mathrm{logb}+\left(\mathrm{p}-\mathrm{q}\right)\mathrm{logc} \\ =\log \left({\mathrm{a}}^{\mathrm{q}-\mathrm{r}}\times {\mathrm{b}}^{\mathrm{r}-\mathrm{p}}\times {\mathrm{c}}^{\mathrm{p}-\mathrm{q}}\right) \end{gathered}$$

a^q-r = x^q-r * y^(p-1)(q-r)

b^r-p = x^r-p * y^(q-1)(r-p)

c^p-q = x^p-q * y^(p-q)(r-1)
multiply all 3 of them, we get RHS,
=x⁰y⁰
= 1
So $\log \left(a^{q-r}\times b^{r-p}\times c^{p-q}\right)=\log 1=0$

Regards