Question

The maximum sum of series $20 + 19\dfrac{1}{3} + 18\dfrac{2}{3} + ...$ is
(A). 310
(B). 300
(C). 320
(D). None of these

Answer 1

Hint – Given series is $20 + 19\dfrac{1}{3} + 18\dfrac{2}{3} + ...$ , now this series can be written in the form $20 + \dfrac{{58}}{3} + \dfrac{{56}}{3} + ...$ by converting into whole fraction. Now, the given series is in arithmetic progression, with the first term is 20, common difference is -2/3.

Common step-by-step solution -
We have been given the series-
$20 + 19\dfrac{1}{3} + 18\dfrac{2}{3} + ...$
The given series can be written as-
$20 + \dfrac{{58}}{3} + \dfrac{{56}}{3} + ...$
Now, as we can see the given series is in A.P. since the difference between any two consecutive terms is same, i.e., $\dfrac{{58}}{3} - 20 = - \dfrac{2}{3}$ and also $\dfrac{{56}}{3} - \dfrac{{58}}{3} = - \dfrac{2}{3}$ .
So, in this case, the first term of the series is, a = 20.
The common difference is $d = \dfrac{{56}}{3} - \dfrac{{58}}{3} = - \dfrac{2}{3}$.
Now, we know in A.P., the terms are given by the formula-
${t_n} = a + (n - 1)d$
Putting the values of a, d in the above equation we get-
${t_n} = 20 + (n - 1)\left( { - \dfrac{2}{3}} \right)$
As the common difference is negative, the terms will become negative after some stage.
So, the sum is maximum if only positive terms are added.
Hence, ${t_n} = 20 + (n - 1)\left( { - \dfrac{2}{3}} \right) \geqslant 0$
Solving further we get-
$
  20 + (n - 1)\left( { - \dfrac{2}{3}} \right) \geqslant 0 \\
  \dfrac{{60 - 2(n - 1)}}{3} \geqslant 0 \\
   \Rightarrow 60 - 2(n - 1) \geqslant 0 \\
   \Rightarrow 60 - 2n + 2 \geqslant 0 \\
   \Rightarrow 62 - 2n \geqslant 0 \\
   \Rightarrow 62 \geqslant 2n \\
   \Rightarrow 31 \geqslant n \\
 $
Now, this implies that the first 31 terms are non-negative.
Now, sum in an A.P. is given by the formula, ${S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$
For maximum sum, we need all positive terms, so n will be 31.
So, the maximum sum is -
\[
  {S_n} = \dfrac{{31}}{2}\left[ {2 \times 20 + (31 - 1) \times \dfrac{{ - 2}}{3}} \right] \\
   = \dfrac{{31}}{2}\left[ {40 + (30) \times \dfrac{{ - 2}}{3}} \right] \\
   = \dfrac{{31}}{2}[40 - 20] = \dfrac{{31}}{2} \times 20 = 310 \\
 \]
Therefore, the maximum sum of the given series is 310.
Hence, the correct option is A. 310.

Note- Whenever such types of questions appear when it is asked to find the maximum sum of the series, then first convert the series into a simpler form, then see whether the series in A.P., if it is in A.P., as mentioned in the solution then find the common difference. As mentioned in the solution the common difference is negative, the terms will become negative after some stage. So, the sum is maximum if only positive terms are added. So, find the number of positive terms which are positive and then find the sum.