Question

Find the sum of the following series-
$7 + 77 + 777 + ...$ to n terms

Answer 1

Hint: Here the given series is not in GP as it does not have a common ratio so first, we will take the number $7$ common from the series. Then multiply and divide each term by $9$. Then, write $9 = \left( {10 - 1} \right)$, $99 = \left( {100 - 1} \right)$, and so on. Then we see that the terms $\left( {10 + 100 + 1000 + ...{\text{n terms}}} \right)$ is in GP as it has the common ratio $10$ so use the formula for the sum of n terms of GP which is given as-
$ \Rightarrow {S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$ Where${\text{r}} > 1$, r= common ratio and a= first term, n= total number of terms

Complete step-by-step answer:
We have to find the sum of given series -$7 + 77 + 777 + ...$ to n terms
Here we see that, $\dfrac{{77}}{7} = 11$ but $\dfrac{{777}}{{77}} = 10.09$
So it is clear that the consecutive terms in this series do not have a common ratio. So this series is not in geometric progression. Then we will make it in GP so we can find the sum of this series easily.
Here on observation, it is clear that the common number in each term is $7$.
So on taking $7$ common from each term, we get-
$ \Rightarrow 7\left\{ {1 + 11 + 111 + ...{\text{n terms}}} \right\}$
Now, multiplying and dividing by the same number in each term will not change the value of the series. So we multiply and divide $9$ in each term.
$ \Rightarrow 7\left\{ {\dfrac{9}{9}\left( {1 + 11 + 111 + ...{\text{n terms}}} \right)} \right\}$
On simplifying, we get-
$ \Rightarrow \dfrac{7}{9}\left\{ {9 + 99 + 999 + ...{\text{n terms}}} \right\}$
We can also write the given equation as-
$ \Rightarrow \dfrac{7}{9}\left\{ {\left( {10 - 1} \right) + \left( {100 - 1} \right) + \left( {1000 - 1} \right) + ...{\text{n terms}}} \right\}$
On separating the terms with unit digit $0$, we get-
$ \Rightarrow \dfrac{7}{9}\left\{ {\left( {10 + 100 + 1000 + ...{\text{n terms}}} \right) - 1 - 1 - 1 - ...{\text{n terms}}} \right\}$
On taking the negative sign common, in the above series, we get-
$ \Rightarrow \dfrac{7}{9}\left\{ {\left( {10 + 100 + 1000 + ...{\text{n terms}}} \right) - \left( {1 + 1 + 1 + ...{\text{n terms}}} \right)} \right\}$
Now, since the digit $1$ is multiplied n times so we can write-
$ \Rightarrow \dfrac{7}{9}\left\{ {\left( {10 + 100 + 1000 + ...{\text{n terms}}} \right) - {\text{n}}} \right\}$ --- (i)
In the above equation, $\left( {10 + 100 + 1000 + ...{\text{n terms}}} \right)$ is in GP as it has the common ratio $\dfrac{{100}}{{10}} = \dfrac{{1000}}{{100}} = 10$
Here the first term ‘a’=$10$ and r=$10$
 Since here ${\text{r}} > 1$ then we use the formula for the sum of n terms of GP which is given as-
$ \Rightarrow {S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$
On putting the given values in this formula, we get-
$ \Rightarrow {S_n} = \dfrac{{10\left( {{{10}^n} - 1} \right)}}{{10 - 1}}$
On solving we get-
$ \Rightarrow {S_n} = \dfrac{{10\left( {{{10}^n} - 1} \right)}}{9}$
Now on substituting this obtained value in eq. (i), we get-
$ \Rightarrow \dfrac{7}{9}\left[ {\dfrac{{10\left( {{{10}^n} - 1} \right)}}{9} - n} \right]$
Since here we do not know the value of n, so we cannot solve it further.
Hence the correct answer is $\dfrac{7}{9}\left[ {\dfrac{{10\left( {{{10}^n} - 1} \right)}}{9} - n} \right]$ .

Note: Here, the formula we have used in the solution is for finite terms of GP. If the common ratio=$1$ then we cannot apply the formula used in the solution. But if the terms are infinite then we use the formula for the sum of infinite terms given by-
$ \Rightarrow {S_\infty } = \dfrac{a}{{1 - r}}$ Where a is the first term, r is the common ratio and $0 < r < 1$