Question

How many terms of the series $1+3+3^2+3^3+\dots$ must be taken to make $3280$.

Answer 1

Step 1: Find the first term and common ratio of the series:

Here, the series is $1+3+3^2+3^3+\dots$ and $S_n=3280$.

First term $a=1$

Common ratio $r=\frac{a_2}{a_1}=\frac{3}{1}=3$

Step 2: Find the value of $n$:

Formula:

Sum of $n$ terms in GP $S_n=\frac{a(r^n-1)}{\left(r-1\right)},r>1$

On substituting the known values,

$\begin{array}{cc}\Rightarrow & \frac{1\left(3^n-1\right)}{\left(3-1\right)}=3280\\ \Rightarrow & 3^{\left(n\right)}-1=3280\times 2\\ \Rightarrow & 3^n=6561\\ \Rightarrow & 3^n=3^8\\ \Rightarrow & n=8\end{array}$

Therefore, need $8$ terms of the series $1+3+3^2+3^3+\dots$ to make $3280$.