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

S seema garhwal

Given :  f (x +y) = f(x) f(y) for all x, y $\epsilon$ N such that  f(1) = 3

$f(1) = 3$

Taking $x=y=1$  , we have

$f(1+1)=f(2)=f(1)*f(1)=3*3=9$

$f(1+1+1)=f(1+2)=f(1)*f(2)=3*9=27$

$f(1+1+1+1)=f(1+3)=f(1)*f(3)=3*27=81$

$f(1),f(2),f(3),f(4).....................$ is $3,9,27,81,..............................$ forms a GP with first term=3 and common ratio = 3.

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

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

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

$40=\frac{(1-3^n)}{-2}$

$-80=(1-3^n)$

$-80-1=(-3^n)$

$-81=(-3^n)$

$3^n=81$

Therefore, $n=4$

Thus, value of n is 4.

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