Question

if the sum of first n terms of series
$20 + 19 \frac{3}{5} + 19 \frac{1}{5} + 18 \frac{4}{5} + ....is \: 488$
and nth term is negative then find n ​

Answer 1

Given series is an AP of first term $a=20$ and common difference $d=-\dfrac{2}{5}.$

Then $n^{th}$ term is,

$\longrightarrow a_n=20-\dfrac{2}{5}(n-1)$

$\longrightarrow a_n=\dfrac{102-2n}{5}$

Given that $n^{th}$ term is negative.

$\longrightarrow a_n<0$

$\longrightarrow\dfrac{102-2n}{5}<0$

$\longrightarrow102-2n<0$

$\longrightarrow2n>102$

$\longrightarrow n>51$

Given that sum of first $n$ terms is 488.

$\longrightarrow\dfrac{n}{2}\,\left(20+a_n\right)=488$

$\longrightarrow n\left(20+\dfrac{102-2n}{5}\right)=976$

$\longrightarrow n\left(202-2n\right)=4880$

$\longrightarrow n\left(101-n\right)=2440$

$\longrightarrow101n-n^2=2440$

$\longrightarrow n^2-101n+2440=0$

$\longrightarrow n^2-61n-40n+2440=0$

$\longrightarrow n(n-61)-40(n-61)=0$

$\longrightarrow (n-61)(n-40)=0$

Since $n>51,$

$\longrightarrow\underline{\underline{n=61}}$

Answer 2

Hey! mate see ur answer in attachment