Question

Find the sum of the series: $\dfrac{2}{9} + \dfrac{1}{3} + \dfrac{1}{2} + ............. + \dfrac{{81}}{{32}}$.

Answer 1

For solving this particular question , you have to check whether it is in G.P or not. If it is in G.P then find the common ratio then the number of the terms present in the series and lastly apply the formula of G.P for finding the sum of the series

Complete step by step solution:
The given series is, $\dfrac{2}{9} + \dfrac{1}{3} + \dfrac{1}{2} + ............. + \dfrac{{81}}{{32}}$.Now, we will check whether this series in G.P or not.For this we have to calculate the ratio, if the ratio of every term with its previous term is constant , then we can say that the series is in G.P.Let us calculate,taking first and second term for evaluating the ratio.We will get the following result,
$\Rightarrow \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{\dfrac{1}{3}}}{{\dfrac{2}{9}}} \\\ \Rightarrow \dfrac{{{a_2}}}{{{a_1}}}= \dfrac{1}{3} \times \dfrac{9}{2} \\\ \Rightarrow \dfrac{{{a_2}}}{{{a_1}}}= \dfrac{3}{2}$

Now, taking second and third term for evaluating the ratio ,
We will get the following result ,
$\Rightarrow \dfrac{{{a_3}}}{{{a_2}}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{1}{3}}}\\\ \Rightarrow \dfrac{{{a_3}}}{{{a_2}}} = \dfrac{1}{2} \times \dfrac{3}{1} \\\ \Rightarrow \dfrac{{{a_3}}}{{{a_2}}} = \dfrac{3}{2}$
We get the common ratio , $r = \dfrac{3}{2}$ .
And the given last term , ${t_n} = \dfrac{{81}}{{32}}$ ,
And we know that ${t_n} = a{r^{n - 1}}$ ,
Therefore ,
$\dfrac{{81}}{{32}} = \dfrac{2}{9} \times {\left( {\dfrac{3}{2}} \right)^{n - 1}} \\\ \Rightarrow \dfrac{{81}}{{32}} = \dfrac{2}{9} \times {\left( {\dfrac{3}{2}} \right)^{ - 1}} \times {\left( {\dfrac{3}{2}} \right)^n} \\\ \Rightarrow \dfrac{{81}}{{32}} \times \dfrac{9}{2} \times \left( {\dfrac{3}{2}} \right) = {\left( {\dfrac{3}{2}} \right)^n} \\\ \Rightarrow {\left( {\dfrac{3}{2}} \right)^7} = {\left( {\dfrac{3}{2}} \right)^n} \\\ \Rightarrow n = 7 \\\$
Now , we know the common ratio , number of terms.Therefore, we can easily calculate the sum of series which in G.P , by using the following formula ,
${S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}$
Where ‘Sn’ is the sum of the series , which we have to find, ‘a’ is the very first term of the given series that is $\dfrac{2}{9}$, ‘r’ is the common ratio , which we had calculated above , $r = \dfrac{3}{2}$ ,
‘n’ is the total number of terms present in the series that is , $n = 7$ .
Now, substitute the above values in the formula , we will get ,
$\Rightarrow {S_n} = \dfrac{{\dfrac{2}{9}\left( {1 - {{\left( {\dfrac{3}{2}} \right)}^7}} \right)}}{{1 - \dfrac{3}{2}}}\\\ \therefore {S_n}= \dfrac{{2059}}{{288}}$

Hence the sum of the series is $\dfrac{{2059}}{{288}}$.

Note: While applying the formula of sum of ‘nth’ term of G.P. we have to make sure that the given series must be in G.P. We can check it by evaluating the ratios. If the ratio of every term with its previous term is constant , then we can say that the series is in G.P.