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

Solution :
$$\eqalign{ & \because f\left( {x + y} \right) = f\left( x \right).f\left( y \right) \cr & \Rightarrow {\text{ Let }}f\left( x \right) = {t^x} \cr & \because f\left( a \right) = 2\therefore t = 2 \cr & \Rightarrow f\left( x \right) = {2^x} \cr & {\text{Since, }}\sum\limits_{k = 1}^{10} f \left( {a + k} \right) = 16\left( {{2^{10}} - 1} \right) \cr & {\text{Then }},\sum\limits_{k = 1}^{10} {{2^{a + k}}} = 16\left( {{2^{10}} - 1} \right) \cr & \Rightarrow {2^a}\sum\limits_{k = 1}^{10} {{2^k}} = 16\left( {{2^{10}} - 1} \right) \cr & \Rightarrow {2^a} \times \frac{{\left( {\left( {{2^{10}}} \right) - 1} \right) \times 2}}{{(2 - 1)}} = 16 \times \left( {{2^{10}} - 1} \right) \cr & \Rightarrow {2.2^a} = 16 \Rightarrow {\text{a}} = 3 \cr} $$