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Q.

If f is a function satisfying f(x+y)=f(x)f(y) for all x,y∈N such that f(1)=3 and $\sum_{x = 1}^{n} f (x) = 120$ then find the value of n.

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a

5

b

4

c

3

d

2

answer is A.

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Detailed Solution

Geometric sequence otherwise known as Geometric progression, abbreviated as GP is a type sequence where the ratio between any two consecutive numbers is constant. If (xn)=x1,x2,x3,...is an GP, then

\frac{x_{2}}{x_{1}} = \frac{x_{3}}{x_{1}} \dots . = r \dots . .

...(1)
Here r is called common ratio and if we take the first term

x_{1} = a

, then sum of GP of first n terms is given by,
S =

\frac{a (r^{n} - 1)}{r - 1}

We are given in the question the summation value

\sum_{x = 1}^{n} f (x) = 120

which we can write by expansion as;
f(1) + f(2)+...+f(n)=120......(1)
We are also given the following functional equation.
f(x+y) = f(x)f(y).........(2)
We are also given f(1)=3. Let us put y=1 in the above equation to have;
f(1+1) = f(1)f(1)
⇒ f(2) = 3⋅3 =

3^{2}

We put x=1,y=2 in equation (2) and use previously obtained f(2)=

3^{2}

to have
f(2+1) = f(1)f(2)
⇒f(3) = 3⋅

3^{2}

⇒f(3) =

3^{3}

We can go on putting y=3,4,...n+1 to have values of the function as
f(3)=

3^{4}

,f(4)=

3^{5}

...,f(n)=

3^{n}

We put these values in equation (1) to have;
3+

3^{2}

+...+

3^{n}

=120
We see that in the left hand side of the above step, we have have GP with first term say a=3 and common ratio r=

\frac{3^{2}}{3}

=

\frac{3^{3}}{3^{2}} \dots \dots \dots

=3. We use formula for sum of first n terms of a GP and have;
⇒

\frac{3 (3^{n} - 1)}{3 - 1}

= 120
⇒

\frac{3}{2} {(3}^{n - 1} - 1) = 120

⇒

3^{n - 1} = 120 \times \frac{2}{3} + 1 = 81

⇒

3^{n - 1} = 3^{4}

We equate the exponents of both sides of the above equations to have;
⇒n−1= 4
∴ n= 5

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