JEE MAIN - Mathematics (2020 - 3rd September Morning Slot - No. 22)
The value of $${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$$ is equal to ______.
Answer
4
Explanation
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
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
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