CAT 2021

Slot 3

QA

Question & Solution

$(\sqrt[7]{10})(\sqrt[7]{10})^{2}...(\sqrt[7]{10})^{n}>999$

$(\sqrt[7]{10})^{1+2+...+n}>999$

$10^{\frac{1+2+...+n}{7}}>999$

For minimum value of n,

$\frac{1+2+...+n}{7}=3$

1 + 2 + ... + n = 21

We can see that if n = 6, 1 + 2 + 3 + ... + 6 = 21.