Question

If A = 1 + r^a+ r^2a +r^3a + .... as and
B = 1+ r^b + r^2b +r^3b + .... as,
then a/b is equal to.............​

Answer 1

EXPLANATION.

$\sf : \implies \: a \: = 1 + {r}^{a} + {r}^{2a} + {r}^{3a} + ....... \\ \\ \sf : \implies \: b \: = \: 1 + {r}^{b} + {r}^{2b} + {r}^{3b} + ....... \\ \\ \sf : \implies \: then \: \frac{a}{b} \: equals \: to \:$

$\sf : \implies \: sum \: of \: infinite \: term \: of \: gp \\ \\ \sf : \implies \: s_{ \infty } \: = \frac{a}{1 - r} \: = |x| < 1$

$\sf : \implies \: a \: = 1 + {r}^{a} + {r}^{2a} + {r}^{3a} + ....... \\ \\ \sf : \implies \: first \: term \: = a \: = 1 \\ \\ \sf : \implies \: common \: ratio \: = \: r \: = \frac{b}{a} = \frac{ {r}^{a} }{1} = {r}^{a} \\ \\ \sf : \implies \: s_{ \infty } \: = \frac{1}{1 - {r}^{a} } \\ \\ \sf : \implies \: a \: = \frac{1}{1 - {r}^{a \: } }$

$\sf : \implies \: 1 - {r}^{a} = \dfrac{1}{a} \\ \\ \sf : \implies \: r {}^{a} = 1 - \frac{1}{a} \\ \\ \sf : \implies \: r {}^{a} = \frac{a - 1}{a} \\ \\ \sf : \implies \: take \: log \: on \: both \: sides \: \\ \\ \sf : \implies \: log( {r}^{a} ) = log( \frac{a - 1}{a} ) \\ \\ \sf : \implies \: a log(r {}^{} ) = log( \frac{a - 1}{a} ) \: \: \: \: ......(1)$

$\sf : \implies \: b \: = 1 + {r}^{b} + {r}^{2b} + {r}^{3b} + ....... \\ \\ \sf : \implies \: first \: term \: = a \: = 1 \\ \\ \sf : \implies \: common \: ratio \: = r \: = \frac{b}{a} = \frac{ {r}^{b} }{1} = {r}^{b} \\ \\ \sf : \implies \: s_{ \infty } \: = \frac{1}{1 - {r}^{b} } \\ \\ \sf : \implies \: b \: = \frac{1}{1 - {r}^{b} }$

$\sf : \implies \: 1 - {r}^{b} = \dfrac{1}{b} \\ \\ \sf : \implies \: {r}^{b} = 1 - \frac{1}{b} \\ \\ \sf : \implies \: r {}^{b} = \frac{b - 1}{b} \\ \\ \sf : \implies \: take \: log \: on \: both \: sides \: \\ \\ \sf : \implies \: log( {r}^{b} ) = log( \frac{b - 1}{b} ) \\ \\ \sf : \implies \: b \: log(r) = log( \frac{b - 1}{b} ) \: \: \: \: ........(2)$

$\sf : \implies \: from \: equation \: (1) \: \: and \: \: (2) \: \: we \: \: get \\ \\ \sf : \implies \: \frac{a log(r) }{b log(r) } = \frac{ log( \dfrac{a - 1}{a} ) }{ log( \dfrac{b - 1}{b} ) } \\ \\ \sf : \implies \: \frac{a}{b} = \frac{ log( \dfrac{a - 1}{a} ) }{ log( \dfrac{b - 1}{b} ) }$