Let Σ _k=1^10f(a+k)=16(2^10-1), where the function f satisfies f(x+y)=f(x)f(y) for all natural numbers x, y and f(1)=2. Then the natural number 'a'…

Given $f(1)=2$f(1)=2 and $f(x+y)=f(x)\cdot f(y)$f(x+y)=f(x)⋅f(y)

Put $x=y=1$x=y=1

$$\Rightarrow f(2)=f(1)\cdot f(1)={2}^{2}$$⇒f(2)=f(1)⋅f(1)=22

Now, put $x=1, y=2$x=1,y=2

$$\Rightarrow f(3)=f(1)\cdot f(2)={2}^{3}$$⇒f(3)=f(1)⋅f(2)=23

Now, put $x=1, y=3$x=1,y=3

$$\Rightarrow f(4)=f(1)\cdot f(3)={2}^{4}$$⇒f(4)=f(1)⋅f(3)=24

Similarly, $f(10)={2}^{10}$f(10)=210

Now $\displaystyle \sum _{k=1}^{10} f(a+k)=16({2}^{10}-1)$k=1∑10​f(a+k)=16(210−1)

$$\Rightarrow \sum _{k=1}^{10}f(a)f(k)=16({2}^{10}-1)$$⇒k=1∑10​f(a)f(k)=16(210−1)

$$\Rightarrow f(a)\sum _{k=1}^{10}f(k)=16({2}^{10}-1)$$⇒f(a)k=1∑10​f(k)=16(210−1)

$$\Rightarrow f(a)(f(1)+f(2)\ldots +f(10))=16({2}^{10}-1)$$⇒f(a)(f(1)+f(2)…+f(10))=16(210−1)

$$\Rightarrow f(a)({2}^{1}+{2}^{2}+{2}^{3}+\ldots +{2}^{10})=16({2}^{10}-1)$$⇒f(a)(21+22+23+…+210)=16(210−1)

Using the sum of $n$n terms of a geometric progression, i.e. $a+ar+a{r}^{2}+a{r}^{3}+...+a{r}^{n-1}=\frac{a({r}^{n}-1)}{r-1},$a+ar+ar2+ar3+...+arn−1=r−1a(rn−1)​, we get

$$f(a)\frac{2({2}^{10}-1)}{2-1}=16({2}^{10}-1)$$f(a)2−12(210−1)​=16(210−1)

$$\Rightarrow f(a)=8={2}^{3}$$⇒f(a)=8=23

Also, we know that $f(3)=8$f(3)=8

$$\Rightarrow f(a)=f(3)$$⇒f(a)=f(3)

$$\Rightarrow a=3.$$⇒a=3.