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...(1)
Here r is called common ratio and if we take the first term , then sum of GP of first n terms is given by,
S =
We are given in the question the summation value 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 =
We put x=1,y=2 in equation (2) and use previously obtained f(2)= to have
f(2+1) = f(1)f(2)
⇒f(3) = 3⋅
⇒f(3) =
We can go on putting y=3,4,...n+1 to have values of the function as
f(3)= ,f(4)= ...,f(n)=
We put these values in equation (1) to have;
3++...+=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===3. We use formula for sum of first n terms of a GP and have;
⇒= 120
⇒
⇒
⇒
We equate the exponents of both sides of the above equations to have;
⇒n−1= 4
∴ n= 5
