Question

answer with solution​

Answer 1

Answer:

GIVEN SERIES -

${(343)}^{0.3} {(343)}^{0.03} {(343)}^{0.003} ........... \infty$

USING PROPERTIES -

${(343)}^{0.3 + 0.03 + 0.003 + ......... \infty }$

SUPPOSE THAT -

${(343)}^{p}$

HERE P IS A INFINITE G.P. SERIES.

$p = 0.3 + 0.03 + 0.003 + ......... \infty$

$p = 3(0.1 + 0.01 + 0.001 + ....... \infty )$

$p = 3( \frac{0.1}{1 - 0.1} )$

$p = 3( \frac{0.1}{0.9} )$

$p = \frac{3}{9} = \frac{1}{3}$

SO, Put the value of P

${(343)}^{ \frac{1}{3} } = 7$

SO , 7 IS YOUR ANSWER.

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