Let  $\sum _{k=1}^{10}f(a+k)=16(2^{10}-1)$, where the function $f\left(x\right)$ satisfies $f\left(x+y\right)=f\left(x\right)f\left(y\right)$ for all natural numbers $x,y$ and $f\left(1\right)=2$ Then the natural number ‘$a$’ is:

 $\because f(x+y)=f\left(x\right).f\left(y\right)\Rightarrow Letf\left(x\right)=t^x$
$\because f\left(1\right)=2\therefore t^1=2\Rightarrow f\left(x\right)=2^x$
Since,  $\sum _{k=1}^{10}f(a+k)=16(2^{10}-1)$
Then,  $\sum _{k=1}^{10}{2}^{a+k}=16(2^{10}-1);\Rightarrow \sum _{k=1}^{10}{2}^{{ }^k}=16(2^{10}-1)$
$\Rightarrow 2^a\times \frac{\left(\right(2^{10})-1)\times 2}{(2-1)}=16\times (2^{10}-1)\Rightarrow 2^{a+1}=16$
$\Rightarrow a=3$