Question

Find the maximum sum of the series:
$\rm 20+ 19 \dfrac{1}{3} + 18 \dfrac{2}{3}+ ...$


​

Answer 1

$\Large\text{\underline{\underline{Explanation}}}$

Step 1.

The A.P. sum attains the maximum value at which the term is non-negative.

So, consider where the single term is non-negative -

$\large\cdots\longrightarrow 20-\dfrac{2}{3}(n-1)\geq0.$

Let's solve the inequality.

$\large\cdots\longrightarrow 60-2(n-1)\geq0$

$\large\cdots\longrightarrow -2n+62\geq0$

$\large\cdots\longrightarrow n\leq31$

So, at the 31st term, the A.P. sum attains its maximum value.

Step 2.

We know that -

$\large\text{ \cdots\longrightarrow\boxed{S=\dfrac{n(a+l)}{2}.} }$

From the given A.P., -

• $\largea=20.$
• $\largel=0.$
• $\largen=31.$

Hence, the maximum value of the sum is -

$\large\cdots\longrightarrow S=\dfrac{31}{2}\times(20+0)$

$\large\cdots\longrightarrow\boxed{S=310.}$

Answer 2

$\bf \underline{\: Provided \: series \: with \: us }:$

$\rm 20+ 19 \dfrac{1}{3} + 18 \dfrac{2}{3}+ ...$

$\bf \underline{\:What\: we\: have \: to\: calculate }:$

$\rm \: Maximum \: sum \: of \: given \: series$

$\bf \underline{\:Step\:by\: step \: explanation\: }:$

$\bigstar \: \small \text{First convert the series into a simpler form}$

$\rm \: The \: given \: series \: we \: can \: be \: written \: as :$

$\rm \implies \: \dfrac{60}{3} + \dfrac{58}{3} + \dfrac{56}{3} + \dfrac{54}{3} + ... \dfrac{2}{3} + 0$

$\bigstar \: \small \text{Check whether the series is in A.P }$$\small\text{or not by finding common difference}$

$\rm \: The \: common \: difference \: is$

$\rm \implies \: d =\dfrac{58}{3} - \dfrac{60}{3} = - \dfrac{2}{3}$

$\rm \star \: Note$

$\rm \: We \: take \: the \: first \: \: term \: of \: the \: series ,$

$\rm \: a = \dfrac{60}{3}$

$\bf \bigstar \: Here$

$\rm \implies \: a + (n-1)d<0$

$\implies \rm\: \dfrac{60}{3} + (n - 1) - \dfrac{2}{3}$

$\rm \implies \: n > 31$

$\bf \bigstar \: Now:$

$\rm \implies \: a \: 31 = a + 30 \: d$

$\rm\implies \: \dfrac{60}{3} \times 31 = \dfrac{60}{3} + 30 + ( - \dfrac{2}{3} )$

$\implies \: 0$

$\rm \: Therefore \: the \: 0 \: is \: last \: term \: of \: given \: series$

$\rm \: Since \: the \: difference \: between \: any \: two$$\rm \: consecutive \: terms \: is \: same \: in \: given \: series$$\rm \: so \: the \: given \: series \: is \: in \: A.P$

$\rm \implies\: d \: = \dfrac{56}{3} - \dfrac{58}{3} = - \dfrac{2}{3}$

$\rm \implies\: d \: = \dfrac{54}{3} - \dfrac{54}{3} = - \dfrac{2}{3}$

$\rm \: As \ in \: A.P. \: the \: terms \: are \: given \: by \: this \: formula$

$\rm{ t n= a+(n1) d}$

$\rm{Where}$

$\rm \star \: {a = \dfrac{60}{3} }$

$\rm \star \: {d = - \dfrac{2}{3} }$

$\rm{The \: term \: will \: become \: negative \: after \: some \: stage}$$\rm{as \: common \: difference \: is \: negative.}$$\rm{So \: sum \: is \: maximum \: if \: positive \: terms \: are \: added.}$

$\rm{By \: putting \: this \: value}$

$\rm{\implies tn \: = \: \dfrac{60}{3} +(n−1) -( \dfrac{2}{3} ) \geqslant 0}$

$\rm{\implies \: \dfrac{60}{3} = (n - 1) \: \dfrac{2}{3} }$

$\rm{\implies \: 62 \geqslant 2n }$

$\rm{\implies \: 31 \geqslant n }$

$\rm \: Hence \: sum \: of \: 31 \: \: term \: is \: maximum \: and \: are \: non - negative \:$

$\sf\therefore\underline{Maximum \: \: sum}$

$\rm \therefore Sn = \dfrac{n}{2} \: (first \: term + last \: term)$

$\rm \implies \dfrac{31}{2} \: ( \dfrac{60}{3} + 0 )$

$\rm \implies \: 31 \times 10$

$\rm \implies \boxed{310}$

$\sf\therefore\underline{Therefore}$

$\rm \therefore \: Maximum \: sum \: of \: given \: series \: is \: 310$