Question

How many consecutive terms starting from the first term of the series
(i) $3+9+27+....$ would sum to $1092$?
(ii) $2+6+18+....$ would sum to $728$?

Answer 1

(i) The given geometric series is $3+9+27+....$ in which the first term is $a_1=3$, the second term is $a_2=9$ and the sum is $S_n=1092$.

We find the common ratio $r$ by dividing the second term by first term as shown below:

$r=\frac{9}{3}=3>1$

We know that the sum of an geometric series with first term $a$ and common ratio $r$ is $S_n=\frac{a(r^n-1)}{r-1}$ if $r>1$

Now, substitute $a=3,r=3$ and $S_n=1092$$ in $S_n=\frac{a(r^n-1)}{r-1}$ as follows:

$S_n=\frac{3[{\left(3\right)}^n-1]}{3-1}\Rightarrow 1092=\frac{3}{2}[{\left(3\right)}^n-1]\Rightarrow 1092\times 2=3[{\left(3\right)}^n-1]\Rightarrow \frac{1092\times 2}{3}=[{\left(3\right)}^n-1]$

$\Rightarrow 3^n-1=728\Rightarrow 3^n=728+1\Rightarrow 3^n=729\Rightarrow 3^n=3^6\Rightarrow n=6$

Hence there are $6$ consecutive terms starting from the first term of the series.

(ii) The given geometric series is $2+6+18+....$ in which the first term is $a_1=2$, the second term is $a_2=6$ and the sum is $S_n=728$.

We find the common ratio $r$ by dividing the second term by first term as shown below:

$r=\frac{6}{2}=3>1$

We know that the sum of an geometric series with first term $a$ and common ratio $r$ is $S_n=\frac{a(r^n-1)}{r-1}$ if $r>1$

Now, substitute $a=2,r=3$ and $S_n=728$$ in $S_n=\frac{a(r^n-1)}{r-1}$ as follows:

$S_n=\frac{2[{\left(3\right)}^n-1]}{3-1}\Rightarrow 728=\frac{2}{2}[{\left(3\right)}^n-1]\Rightarrow 728=[{\left(3\right)}^n-1]\Rightarrow 3^n-1=728\Rightarrow 3^n=728+1\Rightarrow 3^n=729\Rightarrow 3^n=3^6\Rightarrow n=6$

Hence there are $6$ consecutive terms starting from the first term of the series.