Geometric progression

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$Let \sum_{k = 1}^{10} f (a + K) = 16 (2^{10} - 1), where the function f (x) satisfies f (x + y) = f (x) f (y) for all$ $natural numbers x, y and f (1) = 2, then the natural number ' a' is$

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Solution

$f (x + y) = f (x) f (y) \Rightarrow f (x) = 2^{x}$ $(∵ f (1) = 2)$
$\sum_{k = 1}^{10} f (a + k) = f (a + 1) + f (a + 2) + ..... + f (a + 10)$
= $2^{a} (2 + 2^{2} + ....... + 2^{10})$
= $2^{a} (2 (2^{10} - 1))$
= $2^{a + 1} (2^{10} - 1)$

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