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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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If $f$ is a function satisfying $f(x+y)=f(x).f(y)\:\:\forall x,y\in N$ such that $f(1)=3\:\:and\:\:\sum ^{n}_{x=1} f(x)=120$, then find the value of $n$.

$\begin{array}{1 1}3 \\ 4 \\ 5 \\ 6 \end{array} $

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  • Sum of $n$ terms of G.P. $=S_n=a\large\frac{r^n-1}{r-1}$
Given: $f(x+y)=f(x).f(y),$
$\qquad\:\sum ^{n}_{x=1} f(x)=120$ and $f(1)=3$.
when $x=y=1$, $f(1+1)=f(2)=f(1).f(1)=3\times 3=9$
when $x=2\:and\:y=1, f(2+1)=f(3)=f(2).f(1)=9\times 3=27$
Now
Given: $\sum ^{n}_{x=1} f(x)==f(1)+f(2)+f(3)+........f(n)=120$
$\Rightarrow\:3+9+27+........f(n)=120$
This series is a G.P. with $a=3,\:\:r=3\:\:and\:\:S_n=120$
We know that Sum of $n$ terms of G.P. $=S_n=a\large\frac{r^n-1}{r-1}$
$\Rightarrow\:3.\large\frac{3^n-1}{3-1}$$=120$
$\Rightarrow\:3^n-1=40\times 2=80$
$\Rightarrow\:3^n=81=3^4$ $\Rightarrow\:n=4$.
answered Mar 29, 2014 by rvidyagovindarajan_1
 

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