# CAT 2019 Question Paper | Quants Slot 1

###### CAT Previous Year Paper | CAT Quants Questions | Question 33

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Question 33 : Consider a function f(x+y) = f(x) f(y) where x , y are positive integers, and f(1) = 2. If f (a+1) + f (a+2) + ..... + f(a+n) = 16 (2n - 1) then a is equal to. [TITA]

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We know that f(x+y) = f(x) f(y)
Let x = a and y = 1,
f (a + 1) = f(a). f (1)
f(a+1) = f(a). 2
f(a+2) = f(a+1) f(1)
f(a+2) = f(a) x 2 x 2 = f(a) x 22
Similarly, f(a+n) = f(a) x 2n
(a+1) + f (a+2) + ..... + f(a+n) = f(a) {21 + 22 + 23+...+2n}
(a+1) + f (a+2) + ..... + f(a+n) = 2f(a) {1 + 2 + 22 + 23+...+2n-1}, which is an infinite GP series
On solving,
(a+1) + f (a+2) + ..... + f(a+n) = 2f(a) {2n - 1}
We know that, (a+1) + f (a+2) + ..... + f(a+n) = 16 (2n - 1)
So, 2f(a) {2n - 1} = 16 (2n - 1)
2 f(a) = 16
f(a) = 8
We know that, f (1) = 2 , f(2) = 4 and f(3) = 8
So, a = 3

The question is " Consider a function f(x+y) = f(x) f(y) where x , y are positive integers, and f(1) = 2. If f (a+1) + f (a+2) + ..... + f(a+n) = 16 (2n - 1) then a is equal to. [TITA]"

##### Hence, the answer is 3

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