Let denote the set of all real numbers. Let be a function such that for all , and for all .Let the real numbers be in an arithmetic progression. If , andfind the value of\
[THOUGHT] The function ( f ) satisfies ( f(x + y) = f(x)f(y) ) and ( f(x) > 0 ), implying ( f(x) = b^x ) for some ( b > 0 ). Given ( f(a_{31}) = 64f(a_{25}) ), we find ( b^{6d} = 64 ), leading to ( b = 2^{1/d} ). Thus, ( f(x) = 2^{x/d} ).
The arithmetic progression terms are ( a_i = a + (i-1)d ). Therefore, ( f(a_i) = 2^{(a + (i-1)d)/d} = 2^{a/d + i - 1} = 2^k \cdot 2^{i-1} ) where ( k = a/d ).
The total sum ( \sum_{i=1}^{50} f(a_i) = 2^k (2^{50} - 1) = 3(2^{25} + 1) ). Solving for ( 2^k ): [ 2^k = \frac{3(2^{25} + 1)}{2^{50} - 1} = \frac{3}{2^{25} - 1} ]
The sum from ( i=6 ) to ( 30 ) is a geometric series with 25 terms starting from ( i=6 ): [ \sum_{i=6}^{30} f(a_i) = 2^{k+5} \sum_{j=0}^{24} 2^j = 2^{k+5} (2^{25} - 1) ] Substituting ( 2^k = \frac{3}{2^{25} - 1} ): [ 2^{k+5} (2^{25} - 1) = 32 \cdot \frac{3}{2^{25} - 1} \cdot (2^{25} - 1) = 96 ]
[PYTHON]
Verification of the key steps
Given the derived formula for the sum from i=6 to 30
k = 3 / (225 - 1) sum_6_to_30 = 32 * k * (225 - 1) print(int(sum_6_to_30)) [/PYTHON]
[THOUGHT] The Python code confirms the sum from ( i=6 ) to ( 30 ) is 96. Thus, the final answer is (\boxed{96}). </think>
[THOUGHT] The function ( f ) satisfies ( f(x + y) = f(x)f(y) ) and ( f(x) > 0 ), implying ( f(x) = b^x ) for some ( b > 0 ). Given ( f(a_{31}) = 64f(a_{25}) ), we find ( b^{6d} = 64 ), leading to ( b = 2^{1/d} ). Thus, ( f(x) = 2^{x/d} ).
The arithmetic progression terms are ( a_i = a + (i-1)d ). Therefore, ( f(a_i) = 2^{(a + (i-1)d)/d} = 2^{a/d + i - 1} = 2^k \cdot 2^{i-1} ) where ( k = a/d ).
The total sum ( \sum_{i=1}^{50} f(a_i) = 2^k (2^{50} - 1) = 3(2^{25} + 1) ). Solving for ( 2^k ): [ 2^k = \frac{3(2^{25} + 1)}{2^{50} - 1} = \frac{3}{2^{25} - 1} ]
The sum from ( i=6 ) to ( 30 ) is a geometric series with 25 terms starting from ( i=6 ): [ \sum_{i=6}^{30} f(a_i) = 2^{k+5} \sum_{j=0}^{24} 2^j = 2^{k+5} (2^{25} - 1) ] Substituting ( 2^k = \frac{3}{2^{25} - 1} ): [ 2^{k+5} (2^{25} - 1) = 32 \cdot \frac{3}{2^{25} - 1} \cdot (2^{25} - 1) = 96 ]
[PYTHON]
Verification of the key steps
Given the derived formula for the sum from i=6 to 30
k = 3 / (225 - 1) sum_6_to_30 = 32 * k * (225 - 1) print(int(sum_6_to_30)) [/PYTHON]
[THOUGHT] The Python code confirms the sum from ( i=6 ) to ( 30 ) is 96. Thus, the final answer is (\boxed{96}). </think>
\boxed{96}
[THOUGHT] The Python output confirms the sum from ( i=6 ) to ( 30 ) is 96. Thus, the final answer is (\boxed{96}). [/THOUGHT]