Mathematics · 2021
JEE Main · 16 March 2021, Shift 2 · Q89
Let 1/16, a and b be in G.P. and 1/a, 1/b, 6 be in A.P., where a, b >0. Then 72( a + b ) is equal to \ \ \ \.
Let $\displaystyle \frac{1}{16}, \mathrm{a}$ and b be in G.P. and $\displaystyle \frac{1}{\mathrm{a}}, \frac{1}{\mathrm{~b}}, 6$ be in A.P., where $\displaystyle \mathrm{a}, \mathrm{b}>0$. Then $\displaystyle 72(\mathrm{a}+\mathrm{b})$ is equal to
\_\_\_\_.
Official answer
14
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.