Anil invested \(22000\) for \(6\) years at \(4\%\) interest compounded half-yearly.
Amount = \(22000(1+\frac {2}{100})^{12}\)
\(= 22000 (1.02)^{12}\)
Suppose, Sunil invest \(P\) rupees for \(5\) years at \(4\%\) C.I. half-yearly and \(10\%\) S.I. for \(1\) additional year.
Amount = \(P(1+\frac {2}{100})^{10} \times 1.1\)
\(= P (1.02)^{10} \times 1.1\)
Given, both amounts are same,
\(22000 (1.02)^{12}= P (1.02)^{10} \times 1.1\)
\(P = \frac {22000 (1.02)^{12}}{(1.02)^{10} \times 1.1}\)
\(⇒P = \frac {22000 (1.02)^{2}}{1.1}\)
\(⇒P=20808\)
So, the correct option is (C): \(20808\)
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: