Let R denote the set of all real numbers. Let | vedclass

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(B) Given $f(x+y)=f(x)f(y)$ and $f(x)>0$,the function is of the form $f(x)=k^x$ for some $k>0$.Since $f(a_{31})=64 f(a_{25})$,we have $k^{a+30d}=64 k^

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$96$

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(B) Given $f(x+y)=f(x)f(y)$ and $f(x)>0$,the function is of the form $f(x)=k^x$ for some $k>0$.Since $f(a_{31})=64 f(a_{25})$,we have $k^{a+30d}=64 k^{a+24d}$,where $a$ is the first term and $d$ is the common difference.This simplifies to $k^{6d}=64$,so $k^d=2$.The sum $\sum_{i=1}^{50} f(a_i) = k^a + k^{a+d} + \ldots + k^{a+49d} = k^a \frac{(k^d)^{50}-1}{k^d-1} = k^a \frac{2^{50}-1}{2-1} = k^a(2^{50}-1)$.Given $k^a(2^{50}-1) = 3(2^{25}+1)$,we have $k^a(2^{25}-1)(2^{25}+1) = 3(2^{25}+1)$,so $k^a = \frac{3}{2^{25}-1}$.We need to find $\sum_{i=6}^{30} f(a_i) = k^{a+5d} + k^{a+6d} + \ldots + k^{a+29d} = k^{a+5d} \frac{(k^d)^{25}-1}{k^d-1} = k^a (k^d)^5 (2^{25}-1)$.Substituting the values: $\frac{3}{2^{25}-1} \cdot 2^5 \cdot (2^{25}-1) = 3 \cdot 32 = 96$.

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