Question

If a–¹ , b–¹, c–¹ are in A.P (abc ≠ 0), then a-b/b-c is equal to​

Answer 1

$\bold{ {a}^{ - 1} ,\: \: {b}^{ - 1} ,\: \: \: {c}^{ - 1} } \: \: are \: in \: AP.$

$\bold{So, \: \: b^{-1} = \frac{ {a}^{ - 1} + {c}^{ - 1} }{2} }$

$\implies \: { \bold{b \: = \frac{2}{ {a}^{ - 1} + {c}^{ - 1} } }}$

$\implies \: { \bold{b \: = \frac{2}{ \frac{1}{a} + \frac{1}{c} } }}$

$\implies \: { \bold{b \: = \frac{2}{ \frac{c + a}{ac} } }}$

$\implies \: \boxed{ \bold{b \: = \frac{2ac}{ c + a} }}$

Now ,

$\large \bold{ \frac{a - b}{b - c} }$

$= \large \bold{\frac{a - \frac{2ac}{c + a} }{\frac{2ac}{c + a} - c} }$

$= \large \bold{\frac{ \frac{ac - {a}^{2} - 2ac}{c + a} }{\frac{2ac - {c}^{2} - ca }{c + a} } }$

$= \large \bold{{ \frac{ac - {a}^{2} - 2ac}{c + a} } \times {\frac{c + a }{2ac - {c}^{2} - ca } }}$

$= \large \boxed{\bold{ \frac{ - {a}^{2} - ac}{ {c}^{2} + ac} }}$