Question

Given the sequence ${10}^{\frac{1}{11}},{10}^{\frac{2}{11}},{10}^{\frac{3}{11}},......{10}^{\frac{n}{11}}$. The smallest value of $nϵN$ such that the product of the first n terms of the sequence exceeds one lac, is

• A. $9$
• B. $10$
• C. $11$
• D. $12$

Answer 1

The correct option is C $11$

The product of first n terms =${10}^{\frac{1}{11}}\times {10}^{\frac{2}{11}}\times {10}^{\frac{3}{11}}\times .......{10}^{\frac{n}{11}}={10}^{\frac{1}{11}+\frac{2}{11}+\frac{3}{11}+.....+\frac{n}{11}}={10}^{\frac{1+2+3+.....+n}{11}}={10}^{\frac{n(n+1)}{2\times 11}}$
Now, ${10}^{\frac{n(n+1)}{2\times 11}}>1lac$
$\Rightarrow {10}^{\frac{n(n+1)}{2\times 11}}>{10}^5\Rightarrow \frac{n(n+1)}{2\times 11}>5\Rightarrow n^2+n-110>0\Rightarrow (n+11)\times (n-10)>0\Rightarrow n\in (-\infty ,-11)\cup (10,\infty )$
but given that $n\in N$ hence $n\in (10,\infty )$

So smallest $n=11$