Question

Find the number of terms in the series $20 + 19\dfrac{1}{3} + 18\dfrac{2}{3} + ...$ of which the sum is $300$, explain the double answer.

Answer 1

Hint: In this question, the difference between any two consecutive terms of series is equal. So, we can say that the given series is in arithmetic progression. To solve this question, we will use the formula of sum of first $n$ terms of series which is given by ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ where ${S_n}$ is sum of first $n$ terms, $a$ is the first term of the series and $d$ is the difference between any two consecutive terms. In this formula, we will substitute the given sum to find the number of terms.

Complete step-by-step solution:
First we will rewrite the given series. That is, $20 + \dfrac{{58}}{3} + \dfrac{{56}}{3} + .....$. Here the first term is $a = 20$ and the common difference is $d = \dfrac{{58}}{3} - 20 = - \dfrac{2}{3}$. Also given that sum is $300$. That is, ${S_n} = 300$.
Now we are going to use the formula of sum of first $n$ terms of series which is given by ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$. Put values of $a$, $d$ and ${S_n}$ in the formula to find the number of terms.
Therefore, $300 = \dfrac{n}{2}\left[ {2\left( {20} \right) + \left( {n - 1} \right)\left( { - \dfrac{2}{3}} \right)} \right]$
$
   \Rightarrow 600 = n\left[ {40 - \left( {n - 1} \right)\dfrac{2}{3}} \right] \\
   \Rightarrow 600 = \dfrac{n}{3}\left[ {120 - 2\left( {n - 1} \right)} \right] \\
   \Rightarrow 1800 = n\left[ {120 - 2n + 2} \right] \\
   \Rightarrow 1800 = n\left[ {122 - 2n} \right] \\
   \Rightarrow 1800 = 122n - 2{n^2} \\
   \Rightarrow 2{n^2} - 122n + 1800 = 0 \\
$
Now divide by $2$ on both sides,
$
   \Rightarrow {n^2} - 61n + 900 = 0 \\
   \Rightarrow {n^2} - 36n - 25n + 900 = 0 \\
   \Rightarrow n\left( {n - 36} \right) - 25\left( {n - 36} \right) = 0 \\
   \Rightarrow n\left( {n - 36} \right) - 25\left( {n - 36} \right) = 0 \\
   \Rightarrow \left( {n - 36} \right)\left( {n - 25} \right) = 0 \\
$
$ \Rightarrow n = 36$ or $n = 25$

Hence, the number of terms are $36$ or $25$.

Here, the given series is in descending order. So, there is a possibility for negative numbers to occur in the given series. So, we can say that the sum of terms between ${25^{th}}$ and ${36^{th}}$ terms should be zero. So, the sum $300$ is possible for first $36$ terms or for first $25$ terms of the given series.

Note: If the given series is in arithmetic progression and if we know the first term and last term of that series then we can use the formula ${S_n} = \dfrac{n}{2}\left( {a + l} \right)$ to find the sum of first $n$ terms of the series where $a$ is the first term of the series and $l$ is the ${n^{th}}$ term of the series.