QJEEM20Ma4d3f: JEE Main 2020 Mathematics - Class 11 JEEM20Ma4d3f Solution

Given, ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$

As sum of GP upto infinity = ${a \over {1 - r}}$

$\therefore$ ${1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....\infty$ = ${{{1 \over 3}} \over {1 - {1 \over 3}}}$ = ${1 \over 2}$

$\therefore$ ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$

= ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$

= ${\left( {{{16} \over {100}}} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$

= ${\left( {{4 \over {10}}} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$

= ${\left[ {{{\left( {{{10} \over 4}} \right)}^{ - 2}}} \right]^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$

= ${\left[ {{{\left( {2.5} \right)}^{ - 2}}} \right]^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$

= ${{{\left( {2.5} \right)}^{ - 2{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}}$

= ${{{\left( {{1 \over 2}} \right)}^{ - 2}}}$ = 4