Question

25. Ifa^p=b^q=c^r and b^2=ac the value of q(p+r)/pr is given by
(a) 1
(b) -1
(c) 2​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

${a}^{p} = {b}^{q} = {c}^{r} \: \: and \: \: \: {b}^{2} = ac$

TO DETERMINE

$\displaystyle \: \frac{q(p + r)}{pr}$

EVALUATION

${a}^{p} = {b}^{q} = {c}^{r} = k(say)$

$\displaystyle \: \implies \: a = {k}^{ \frac{1}{p} } \: , \: b = {k}^{ \frac{1}{q} } \: , \: \: c = {k}^{ \frac{1}{r} }$

Now

${b}^{2} = ac$

$\displaystyle \implies \: \: \ { \bigg({k}^{ \frac{1}{q} }\bigg)}^{2} = {k}^{ \frac{1}{p} } \times \: {k}^{ \frac{1}{r} }$

$\displaystyle \: \implies \: \: \ {k}^{ \frac{2}{q} } = {k}^{ \: ( \frac{1}{p} + { \frac{1}{r} }) }$

$\displaystyle \: \implies \: \: \ { \frac{2}{q} } = { \: \frac{1}{p} + { \frac{1}{r} } }$

$\displaystyle \: \implies \: \: \ { \frac{2}{q} } = { \: \frac{p + r}{pr} }$

$\implies \: \displaystyle \: \frac{q(p + r)}{pr} = 2$

RESULT

$\sf{THE \: REQUIRED \: ANSWER \: IS \: = 2 \: }$