Question

The sum of 100 terms of the series \[0.9 + 0.09 + 0.009 + .......\] will be:
1) $1 - {\left( {\dfrac{1}{{10}}} \right)^{100}}$
2) $1 + {\left( {\dfrac{1}{{10}}} \right)^{100}}$
3) $1 + {\left( {\dfrac{1}{{100}}} \right)^{100}}$
4) $1 - {\left( {\dfrac{1}{{100}}} \right)^{100}}$

Answer 1

Hint: Consider the given series and find out the common ratio to check whether it is geometric progression or else find the common difference to check whether it is arithmetic progression. As the given series is a geometric series thus, apply the formula of finite sum of geometric series, ${S_n} = \dfrac{{{a_1}\left( {1 - {r^n}} \right)}}{{1 - r}},r \ne 1$ to find the sum of 100 terms.

Complete step by step solution:
Consider the given series as \[0.9 + 0.09 + 0.009 + \ldots \ldots \]up to 100 terms
We will find the common ratio by dividing the second term by the first term, third term by the second term and so on.
$
   \Rightarrow {r_1} = \dfrac{{0.09}}{{0.9}} = 0.1 \\
   \Rightarrow {r_2} = \dfrac{{0.009}}{{0.09}} = 0.1 \\
$
As the common ratio is the same so it is a geometric progression (G.P.) series.
We use the formula of the sum of ${n^{th}}$ terms of G.P given by:
${S_n} = \dfrac{{{a_1}\left( {1 - {r^n}} \right)}}{{1 - r}},r \ne 1$
Where, \[n\] is the number of terms, \[{a_1}\] is the first term and \[r\] is the common ratio.
For the given series,
\[
  {\text{ }}n = {\text{100}} \\
  \;{a_1} = {\text{ 0.9}} \\
  r\; = {\text{ 0.1}} \\
\]
Now, we will substitute the above values in the formula of sum of G.P. series ${S_n} = \dfrac{{{a_1}\left( {1 - {r^n}} \right)}}{{1 - r}},r \ne 1$.
Thus, we have,
$
   \Rightarrow {S_{100}} = \dfrac{{0.9\left( {1 - {{0.1}^{100}}} \right)}}{{1 - 0.1}} \\
   \Rightarrow {S_{100}} = \dfrac{{0.9\left( {1 - {{0.1}^{100}}} \right)}}{{0.9}} \\
   \Rightarrow {S_{100}} = 1 - {\left( {\dfrac{1}{{10}}} \right)^{100}} \\
$
Hence option A i.e. $1 - {\left( {\dfrac{1}{{10}}} \right)^{100}}$ is the required answer of this question.

Note: We have used the formula ${S_n} = \dfrac{{{a_1}\left( {1 - {r^n}} \right)}}{{1 - r}},r \ne 1$ for finding the sum of finite geometric series because $r \leqslant 1$, if $r > 1$ then we cannot use this formula to determine the sum of finite geometric series. As the common ratio is same thus, we have further used the sum formula for geometric series, if it is not same then we might have used the arithmetic series method.