Let $R$ denote the set of all real numbers. Let $f : R \to R$ be a function such that $f(x) > 0$ for all $x \in R$, and $f(x + y) = f(x) f(y)$ for all $x, y \in R$. Let the real numbers $a_1, a_2, \ldots a_{50}$ be in an arithmetic progression. If $f(a_{31}) = 64 f(a_{25})$, and $\sum_{i=1}^{50} f(a_i) = 3(2^{25} + 1)$, then the value of $\sum_{i=6}^{30} f(a_i)$, is

Let R denote the set of all real numbers. Let f : R → R be a function such that f(x) > 0 for all x ∈ R, and f(x + y) = f(x) f(y) for all x, y ∈ R. Let the real numbers a₁, a₂, … a₅₀ be in an arithmetic progression. If f(a₃₁) = 64 f(a₂₅), and ∑ᵢ₌₁⁵⁰ f(aᵢ) = 3(2²⁵ + 1), then the value of ∑ᵢ₌₆³⁰ f(aᵢ), is

$\text{Given: } f(x)>0,\quad f(x+y)=f(x)\,f(y)$

$f(a_{31}) = 64\,f(a_{25})$

$\sum_{i=1}^{50} f(a_i) = 3\left(2^{25}+1\right)$

$\text{To find: } \displaystyle\sum_{i=6}^{30} f(a_i)$

$\text{1. Positivity + multiplicativity } \Rightarrow f(a_i)\ \text{is a GP}$

$\text{2. Ratio } f(a_{31})/f(a_{25}) = r^{6} \Rightarrow r$

$\text{3. Use the 50-term sum } \Rightarrow A = f(a_1)\text{, then the asked sum}$

$a_i = a_1 + (i-1)d$

$f(a_i) = f(a_1)\,[f(d)]^{\,i-1}$

$f(a_i) = A\,r^{\,i-1},\quad A = f(a_1),\ r = f(d) > 0$

$\dfrac{f(a_{31})}{f(a_{25})} = \dfrac{A r^{30}}{A r^{24}} = r^{6}$

$r^{6} = 64 = 2^{6} \;\Rightarrow\; r = 2$

$\sum_{i=1}^{50} f(a_i) = \sum_{i=1}^{50} A r^{\,i-1} = \dfrac{A(r^{50}-1)}{r-1}$

$= A(2^{50}-1)$

$A(2^{50}-1) = 3(2^{25}+1)$

$2^{50}-1 = (2^{25}-1)(2^{25}+1)$

$A = \dfrac{3}{2^{25}-1}$

$\sum_{i=1}^{25} f(a_i) = A(2^{25}-1) = 3$

$\sum_{i=26}^{50} f(a_i) = 2^{25}\sum_{i=1}^{25} f(a_i) = 3\cdot 2^{25}$

$\sum_{i=1}^{50} f(a_i) = 3 + 3\cdot 2^{25} = 3(2^{25}+1)\ \checkmark$

$\sum_{i=6}^{30} f(a_i) = A r^{5}\,\dfrac{r^{25}-1}{r-1}$

$= A\cdot 2^{5}\,(2^{25}-1) = 32\,A(2^{25}-1)$

$\sum_{i=6}^{30} f(a_i) = 32\cdot\dfrac{3}{2^{25}-1}\cdot(2^{25}-1)$

$= 32\cdot 3 = 96$